Physics · Secondary School · Electrostatics

Electrostatics

The shock from a door handle on a dry day, the balloon that sticks to the wall and the lightning in a storm all come from electric charges at rest, or nearly so. We start from charge and the ways of charging an object, measure the Coulomb force, describe the field and the potential that charges create and finish with capacitors, which store electrical energy. In every calculation we use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\) and \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

1Electric charge
2Charging
3Coulomb's law
4Electric field
5Potential
6Plates and conductors
7Capacitors
✓Challenges

STEP 1

Where does electric charge come from?

Matter is made of atoms, with protons and neutrons in the nucleus and electrons around it. The proton and the electron have charges of the same size and opposite signs, and we call that size the elementary charge, \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

A neutral object has as many electrons as protons. In everyday processes it becomes charged by gaining or losing electrons, because the protons stay locked in the nuclei, and so the charge of any object is a whole-number multiple of \(e\). We say that charge is quantised.

Charge is also conserved. When we rub two objects together, the electrons one loses the other gains, and the sum of the charges stays the same as before; no experiment so far has shown net charge being created or destroyed.

In metals, some electrons from each atom move freely through the material, and a charge placed at one point soon spreads out; we call these materials conductors. In plastics, glass and rubber, which we call insulators, the electrons stay bound to their atoms, and the charge tends to stay where we put it.

\(Q = n\,e\)\(e = 1.6 \cdot 10^{-19}\ \text{C}\)\(Q_{\text{before}} = Q_{\text{after}}\)\(n\) is the number of electrons in excess or missing. In the simulation, each blue dot is an electron and each circle with a + is an atom; in a real object, these numbers run to billions of billions. The motion of the electrons follows a simplified model, in two dimensions.

Let's discuss

With the metal, add three electrons at the left end and watch the slices under the object. Where does the extra charge go?
Repeat with the plastic. The charge seems to stay put at the end; why does that happen now?
Remove electrons until the net charge reaches +5e. How many protons and how many electrons does the object have at that moment?
Try to reach a charge of half an \(e\). Does the simulation allow it? Would nature allow it?

Protons

Electrons

Excess or shortfall

Net charge

In units of e

Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
An object receives \(5 \cdot 10^{12}\) more electrons than it had when it was neutral. What is its charge? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
The charge is the number of extra electrons times the elementary charge, with a negative sign, \(Q = -n\,e\) \(= -5 \cdot 10^{12} \cdot 1.6 \cdot 10^{-19}\).
\(Q = -8 \cdot 10^{-7}\ \text{C}\), or \(-0.8\ \mu\text{C}\)
basic
A neutral metal sphere ends up with a charge of \(+3.2\ \mu\text{C}\). Did it gain or lose electrons, and how many? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
A positive charge on an object that was neutral indicates missing electrons, since the protons stay locked in the nuclei.
The number of electrons is \(n = \dfrac{Q}{e} = \dfrac{3.2 \cdot 10^{-6}}{1.6 \cdot 10^{-19}}\).
The sphere lost \(2 \cdot 10^{13}\) electrons.
basic
A plastic ruler and a metal rod are rubbed with a woollen cloth, each held directly in the hand. The ruler becomes charged and attracts bits of paper, and the metal rod seems not to. Why?

Show solution
In the plastic, which is an insulator, the electrons transferred by friction stay stuck where they arrived.
In the metal, which is a conductor, the charge spreads through the rod and drains away through the hand and the body, which also conduct, to the ground.
The metal rod also becomes charged, but loses the charge through our body; held by an insulating handle, it would stay charged.
basic
A classmate claims to have measured a charge of \(2.4 \cdot 10^{-19}\ \text{C}\) on an object. Can the measurement be right? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
The charge of any object is a whole-number multiple of the elementary charge, \(Q = n\,e\).
Here, \(n = \dfrac{2.4 \cdot 10^{-19}}{1.6 \cdot 10^{-19}} = 1.5\), which is not a whole number.
The measurement is wrong, because charge is quantised and there is no such thing as half an electron.
intermediate
Two neutral objects are rubbed against each other, and one of them ends up with a charge of \(-4\ \text{nC}\). What is the charge of the other, and how many electrons passed from one to the other? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
Charge is conserved, and the sum of the two, which was zero, is still zero.
The number of electrons transferred is \(n = \dfrac{4 \cdot 10^{-9}}{1.6 \cdot 10^{-19}}\).
\(+4\ \text{nC}\), with \(2.5 \cdot 10^{10}\) electrons passing to the negative object.
intermediate
In an experiment similar to Millikan's, the charges of four oil drops were measured: \(3.2 \cdot 10^{-19}\), \(4.8 \cdot 10^{-19}\), \(5.6 \cdot 10^{-19}\) and \(8.0 \cdot 10^{-19}\ \text{C}\). Which of the measurements must be wrong? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
Dividing each charge by \(e\), we get 2, 3, 3.5 and 5.
Only whole numbers are possible, because each drop has a whole number of electrons too many or too few.
The measurement of \(5.6 \cdot 10^{-19}\ \text{C}\) must be wrong.
intermediate
One gram of hydrogen has about \(6 \cdot 10^{23}\) protons. What is the total charge of these protons? Why, then, does a gram of hydrogen not exert enormous electric forces on what is around it? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
The charge of the protons is \(Q = n\,e = 6 \cdot 10^{23} \cdot 1.6 \cdot 10^{-19}\), a gigantic value.
The gas, however, has the same number of electrons, with the same negative charge, and the forces from the two parts cancel almost exactly.
The protons add up to \(9.6 \cdot 10^4\ \text{C}\), balanced by the electrons, and the hydrogen is neutral.
intermediate
A charged metal sphere is mounted on a glass stand and keeps its charge for hours. When someone touches it with a finger, the charge disappears almost at once. Explain the two behaviours.

Show solution
Glass is an insulator and stops the charges on the sphere from draining away through the stand.
The human body, in contact with the ground, conducts, and the sphere plus the body and the Earth form an enormous conductor, over which the charge spreads.
The insulating stand holds the charge; the finger earths the sphere, and the Earth takes or supplies electrons until the sphere is practically neutral.
challenge
A 10 g plastic comb has about \(3 \cdot 10^{24}\) electrons. After being run through dry hair, it has \(-8\ \text{nC}\). How many electrons did it gain, and what fraction of the total is that? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
The number of electrons gained is \(n = \dfrac{8 \cdot 10^{-9}}{1.6 \cdot 10^{-19}}\), and the fraction is this number divided by \(3 \cdot 10^{24}\).
\(5 \cdot 10^{10}\) electrons, about \(1.7 \cdot 10^{-14}\) of the total; a tiny imbalance is already enough to attract bits of paper.
challenge
Protons and neutrons are made of three quarks, of two kinds: the u quark, with charge \(+\tfrac{2}{3}\,e\), and the d quark, with charge \(-\tfrac{1}{3}\,e\). The proton is made of uud and the neutron of udd. Work out the charge of each. Does this contradict the quantisation of charge?

Show solution
In the proton, \(\tfrac{2}{3} + \tfrac{2}{3} - \tfrac{1}{3} = 1\), and in the neutron, \(\tfrac{2}{3} - \tfrac{1}{3} - \tfrac{1}{3} = 0\), in units of \(e\).
Quarks have never been observed on their own, only in groups whose total charge is a whole-number multiple of \(e\).
The proton has \(+e\) and the neutron zero; the charge of free particles is still quantised in multiples of \(e\).

STEP 2

How does an object become charged?

In charging by friction, we rub two different materials together, and electrons pass from one to the other. The one that loses electrons becomes positive and the one that gains them becomes negative, with charges of the same size. The triboelectric series ranks materials by how easily they give up electrons, and the comb we run through dry hair, for example, tends to come out negative.

In charging by contact, we touch a charged conductor against another, and the charge is shared out between the two. If the spheres are identical, each ends up with the average of the initial charges, and the total is conserved.

Induction needs no touching. A charged rod near a conductor attracts the charges of the opposite sign to its own and repels those of the same sign, which separate inside the metal. If we earth the conductor at that moment, electrons come up from the ground or go down into it; if the wire is cut before we take the rod away, the conductor is left with a charge of the opposite sign to the rod's.

The balloon rubbed on hair sticks to the wall through a similar effect, polarisation. We can picture the molecules of the wall lining up, with the part of opposite sign a little closer to the balloon, and the attraction, being closer, beats the repulsion. The gold-leaf electroscope makes use of the same separation of charges, with two light metal leaves that spread apart when they carry charges of the same sign.

\(Q_A^{\prime} = Q_B^{\prime} = \dfrac{Q_A + Q_B}{2}\)\(Q_A + Q_B = \text{constant}\)The first relation holds for identical conducting spheres in contact, and the second as long as neither of them exchanges charge with the Earth. In the simulation, the charges are in nanocoulombs (\(1\ \text{nC} = 10^{-9}\ \text{C}\)), the rod has 10 nC and the earth wire is connected to sphere A. The charge induced by the rod, 60% of its own, is an illustrative value, because in practice it depends on the distance and the shape of the objects.

Let's discuss

With A at +6 nC and B at −2 nC, touch the spheres together. How much does each end up with? Did the sum change?
Start again with both spheres neutral, touch them together, bring the rod near and separate them before taking it away. What charge does each one carry off?
With the spheres apart and A neutral, let's repeat the classic induction. Bring the rod near, connect the earth wire, disconnect it and only then take the rod away. With what sign does A end up? And if you take the rod away before disconnecting the earth?
Swap the positive rod for the negative one and repeat the induction. Which way do the electrons move in the earth wire now?

Charge on A

Charge on B

Charge on A

Charge on B

Sum A + B

Charge gained by the Earth

Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A metal sphere A, with a charge of \(+8\ \text{nC}\), touches another identical sphere B, which is neutral, and then the two are separated. What charge does each end up with?

Show solution
Identical spheres in contact share the total charge equally, and the total, \(8 + 0\), is conserved.
Each sphere ends up with \(+4\ \text{nC}\).
basic
In a triboelectric series, glass comes before silk, which means it gives up electrons more easily. If we rub a glass rod with a silk cloth, what sign does each one end up with?

Show solution
The glass loses electrons to the silk during the rubbing.
The one that loses electrons becomes positive and the one that gains them becomes negative, with charges of the same size.
The glass becomes positive and the silk negative.
basic
A positive rod is brought near a neutral metal sphere on an insulating stand. With the rod held still, someone connects the sphere to the Earth with a wire, disconnects the wire and, finally, takes the rod away. With what sign does the sphere end up?

Show solution
The positive rod attracts electrons to the side of the sphere closest to it. Once the earth wire is connected, electrons come up from the ground to the sphere, attracted by the rod.
With the wire disconnected before we take the rod away, these electrons have no way back.
The sphere ends up negative, with the opposite sign to the rod's.
basic
A charged rod is brought near the cap of a neutral gold-leaf electroscope, without touching it. The leaves spread apart. Why, if the electroscope is still neutral?

Show solution
The rod induces a separation of charges in the metal of the electroscope, with charges of the opposite sign to its own on the cap and of the same sign on the leaves, which are far away.
The two leaves end up with charges of the same sign and repel each other.
The total charge is still zero, but induction leaves the leaves with like charges, and they move apart.
intermediate
Three identical metal spheres have charges \(Q_A = +12\ \text{nC}\), \(Q_B = -4\ \text{nC}\) and \(Q_C = 0\). We touch A to B and separate them; then we touch B to C and separate them. What is the final charge on each sphere?

Show solution
In the first contact, A and B end up with the average, \(\dfrac{12 - 4}{2}\) each.
In the second, B and C split in half the charge that B brought from the first contact.
The final sum, \(4 + 2 + 2\), equals the initial one, \(12 - 4 + 0\).
\(Q_A = +4\ \text{nC}\), \(Q_B = +2\ \text{nC}\) and \(Q_C = +2\ \text{nC}\)
intermediate
A party balloon rubbed on hair sticks to a neutral wall. Explain why a charged object attracts a neutral object.

Show solution
The charged balloon polarises the molecules of the wall, and charges of the opposite sign to its own end up a little closer to the balloon than those of the same sign.
Since the electric force decreases with distance, the attraction of the nearer charges beats the repulsion of the more distant ones.
The wall is still neutral, but polarisation leaves a net attraction, which holds the balloon in place.
intermediate
After combing dry hair, the comb becomes negative and some hairs stand on end, apart from one another. What is the sign of the charge on the hairs, and why do they move apart?

Show solution
The comb gained electrons from the hair, and charge is conserved, so the hair lost the same number of electrons.
Hairs with charges of the same sign repel each other.
The hairs become positive and move apart because they repel one another.
intermediate
Two identical metal spheres, with \(+6\ \text{nC}\) and \(-10\ \text{nC}\), are put in contact. What is the final charge on each? How many electrons passed from one to the other, and in which direction? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
Each sphere ends up with the average, \(\dfrac{6 - 10}{2}\).
The sphere that had \(-10\ \text{nC}\) became 8 nC less negative, and lost the equivalent of 8 nC in electrons, \(n = \dfrac{8 \cdot 10^{-9}}{1.6 \cdot 10^{-19}}\).
Both end up with \(-2\ \text{nC}\), and \(5 \cdot 10^{10}\) electrons pass from the negative sphere to the positive one.
challenge
A metal sphere with \(+16\ \text{nC}\) touches, one at a time, three spheres identical to it and initially neutral. What is the final charge on the first sphere, and on each of the other three?

Show solution
At each contact with an identical neutral sphere, the charge on the first is halved, and the sphere it touches takes the other half.
After three contacts, the first is left with \(\dfrac{16}{2^3}\), and the sum of the four charges must still equal the initial 16 nC.
The first ends up with \(+2\ \text{nC}\), and the others with \(+8\), \(+4\) and \(+2\ \text{nC}\).
challenge
Two neutral, identical metal spheres, A and B, are touching each other on insulating stands. A negative rod is brought near A. With the rod still there, the spheres are separated, and only then is the rod taken away. What is the sign of each sphere? What happens if the spheres are then touched together again?

Show solution
The negative rod repels electrons from A to B, which is on the far side. With the spheres separated while the rod is present, A is left short of electrons and B with an excess, in equal amounts.
Touched together again, the identical spheres share the total charge, which is zero.
A becomes positive and B negative, with charges of the same size; touched together again, both become neutral once more.

STEP 3

How hard do two charges push on each other?

Charges of the same sign repel and charges of opposite signs attract. In 1785, Charles Coulomb measured this force with a torsion balance and showed that it is proportional to the product of the charges and inversely proportional to the square of the distance between them.

We apply Coulomb's law to point charges, or to spheres that are small compared with the distance between them. The two charges feel forces of the same size and in opposite directions, as Newton's third law requires, even when one charge is much larger than the other.

The \(1/d^2\) dependence has consequences worth knowing by heart. If we double the distance, the force drops to a quarter; if we halve it, the force becomes four times as large.

The law has the same form as Newton's law of gravitation, and the comparison is striking. Between a proton and an electron, the electric attraction exceeds the gravitational one by a factor of about \(10^{39}\). Gravity dominates on the scale of planets because large bodies are almost always neutral, and the electric forces of their positive and negative charges cancel out.

\(F = k\,\dfrac{|q_1\,q_2|}{d^2}\)\(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\)\(F_g = G\,\dfrac{m_1\,m_2}{d^2}\)We use the value of \(k\) for a vacuum, which hardly changes in air. The charges in the simulation are in microcoulombs (\(1\ \mu\text{C} = 10^{-6}\ \text{C}\)), and the force arrows follow a compressed scale, so that they fit on the screen. The equivalent weight is the mass whose weight, with \(g = 10\ \text{m/s}^2\), would equal the electric force.

Let's discuss

Drag one of the charges and follow the dot on the curve. What happens to the force when the distance is halved?
Use the buttons that double and halve the distance and check the F/4 and 4F marks on the graph.
Change the sign of \(q_2\). Does the size of the force change? And the direction of the arrows?
Set \(q_1 = 10\ \mu\text{C}\) and \(q_2 = 1\ \mu\text{C}\). Are the arrows still the same length on both charges? Why?

Charge q₁

Charge q₂

Distance

Force on each charge

Type

Equivalent weight

Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Two point charges of \(2\ \mu\text{C}\) are 30 cm apart, in a vacuum. What is the size of the force between them? Is it attraction or repulsion? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
By Coulomb's law, \(F = k\,\dfrac{|q_1\,q_2|}{d^2}\) \(= 9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-6} \cdot 2 \cdot 10^{-6}}{0.3^2}\).
The charges have the same sign and repel each other.
\(F = 0.4\ \text{N}\), repulsive
basic
Two charges repel each other with a force of 36 N. What does the force become if the distance between them doubles? And if it triples?

Show solution
The force is inversely proportional to the square of the distance. With double the distance, it is divided by \(2^2 = 4\); with triple, by \(3^2 = 9\).
9 N at double the distance and 4 N at triple.
basic
A charge of \(+3\ \mu\text{C}\) is near another of \(-5\ \mu\text{C}\). Are the forces between them attractive or repulsive? Which of the two charges feels the larger force?

Show solution
Opposite signs attract.
By Newton's third law, the two forces form an action–reaction pair, with the same size and opposite directions, however different the charges may be.
The forces are attractive and have the same size on both charges.
basic
Two charges, of \(1\ \mu\text{C}\) and \(4\ \mu\text{C}\), repel each other with a force of 3.6 N. How far apart are they? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
Isolating the distance in \(F = k\,|q_1\,q_2|/d^2\), we have \(d^2 = \dfrac{k\,q_1\,q_2}{F}\) \(= \dfrac{9 \cdot 10^9 \cdot 4 \cdot 10^{-12}}{3.6}\) \(= 0.01\ \text{m}^2\).
\(d = 0.1\ \text{m}\), or 10 cm
intermediate
Two point charges attract each other with a force \(F\). One of the charges doubles in value and the distance between them is halved. What does the force become?

Show solution
Doubling one of the charges doubles the force, and halving the distance multiplies it by \(2^2 = 4\).
The two effects multiply, \(2 \cdot 4\).
The force becomes \(8F\).
intermediate
In the hydrogen atom, the electron and the proton are about \(5.3 \cdot 10^{-11}\ \text{m}\) apart. Work out the electric force and the gravitational force between them and the ratio of the two. Use \(G = 6.67 \cdot 10^{-11}\ \text{N} \cdot \text{m}^2/\text{kg}^2\), \(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\), \(m_p = 1.67 \cdot 10^{-27}\ \text{kg}\), as well as \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\) and \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
The electric force is \(F_e = \dfrac{k\,e^2}{d^2}\) \(= \dfrac{9 \cdot 10^9 \cdot (1.6 \cdot 10^{-19})^2}{(5.3 \cdot 10^{-11})^2}\) \(\approx 8.2 \cdot 10^{-8}\ \text{N}\).
For the gravitational force, the product \(G\,m_e\,m_p\) is \(6.67 \cdot 10^{-11} \cdot 9.1 \cdot 10^{-31}\) \(\cdot\, 1.67 \cdot 10^{-27} \approx 1.0 \cdot 10^{-67}\), and \(F_g = \dfrac{1.0 \cdot 10^{-67}}{(5.3 \cdot 10^{-11})^2}\) \(\approx 3.6 \cdot 10^{-47}\ \text{N}\).
The ratio is \(\dfrac{F_e}{F_g} = \dfrac{8.2 \cdot 10^{-8}}{3.6 \cdot 10^{-47}}\).
The electric force is about \(2 \cdot 10^{39}\) times the gravitational force.
intermediate
Two small, identical metal spheres, with \(+6\ \mu\text{C}\) and \(-2\ \mu\text{C}\), are 20 cm apart. Work out the force between them. The spheres are then touched together and put back in the same position. What does the force become? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
Before, \(F = 9 \cdot 10^9 \cdot \dfrac{6 \cdot 10^{-6} \cdot 2 \cdot 10^{-6}}{0.2^2}\), attractive.
After the contact, each sphere has \(\dfrac{6 - 2}{2} = 2\ \mu\text{C}\), and \(F^{\prime} = 9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-6} \cdot 2 \cdot 10^{-6}}{0.2^2}\).
2.7 N of attraction before the contact and 0.9 N of repulsion after.
intermediate
Three charges are in a line: \(+2\ \mu\text{C}\) at \(x = 0\), \(+1\ \mu\text{C}\) at \(x = 10\ \text{cm}\) and \(+2\ \mu\text{C}\) at \(x = 30\ \text{cm}\). What is the resultant force on the middle charge? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
The charge on the left pushes the middle one to the right with \(F_1 = 9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-6} \cdot 1 \cdot 10^{-6}}{0.1^2}\) \(= 1.8\ \text{N}\).
The one on the right pushes it to the left with \(F_2 = 9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-12}}{0.2^2}\) \(= 0.45\ \text{N}\).
The resultant is the difference, \(1.8 - 0.45\).
1.35 N, to the right
challenge
A charge of \(+4\ \mu\text{C}\) is at \(x = 0\) and another of \(+1\ \mu\text{C}\) at \(x = 30\ \text{cm}\). At what point between them would a third charge be in equilibrium? Does the answer depend on the value of the third charge?

Show solution
At the point we are looking for, the two forces on the third charge have the same size, \(\dfrac{k \cdot 4\,q}{x^2} = \dfrac{k \cdot 1\,q}{(0.3 - x)^2}\). The values of \(q\) and \(k\) cancel.
Taking the square root, \(\dfrac{2}{x} = \dfrac{1}{0.3 - x}\), and so \(0.6 - 2x = x\).
At \(x = 20\ \text{cm}\), closer to the smaller charge, whatever the third charge is.
challenge
A charged bead is fixed to a table. Another bead, of 1 g and with the same charge, hovers in the air 10 cm above the first, held up by the electric repulsion alone. What is the charge on each bead? Use \(g = 10\ \text{m/s}^2\) and \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
In equilibrium, the electric force balances the weight, \(k\,\dfrac{q^2}{d^2} = m\,g\), with \(m\,g = 10^{-3} \cdot 10 = 10^{-2}\ \text{N}\).
Isolating the charge, \(q = d\,\sqrt{\dfrac{m\,g}{k}}\) \(= 0.1 \cdot \sqrt{\dfrac{10^{-2}}{9 \cdot 10^9}}\).
\(q \approx 1.1 \cdot 10^{-7}\ \text{C}\), about \(0.1\ \mu\text{C}\)

STEP 4

What does a charge do to the space around it?

We can think of a charge as changing the space around it, and of another charge, placed there, as feeling that change. We call the force per unit charge that a small positive test charge would feel at a point the electric field, \(\vec E = \vec F/q\), measured in N/C.

For a point charge \(Q\), the size of the field is \(k|Q|/d^2\), and it does not depend on the test charge. The vector points outwards when \(Q\) is positive and inwards when it is negative, and a negative charge placed at the point feels a force in the direction opposite to the field.

\(\vec E = \dfrac{\vec F}{q}\)\(E = k\,\dfrac{|Q|}{d^2}\)\(\vec E = \vec E_1 + \vec E_2\)The screen represents a region 80 cm wide, with the charges in microcoulombs and \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\). The test charge at P is 1 nC. The length of the orange arrow grows with the logarithm of \(E\), because the field varies a great deal from one point to another, and the dashed arrows are the contributions of each charge.

We draw field lines to see this vector in space, and this may well be the most useful picture of the field. They leave positive charges and arrive at negative ones, the vector \(\vec E\) is tangent to them at each point, and where they are closer together the field is stronger. Two lines do not cross, because the field has only one direction at each point.

With several charges, we use superposition. Each charge creates its field as if it were on its own, and the total field is the vector sum of all of them, which explains, for example, the point of zero field between two equal charges.

Let's discuss

With a single charge, move P far from it and then close to it. How do the size of the arrow and the reading of E follow the distance?
Choose the dipole and move P along a field line. Does the arrow stay tangent to the line?
With two equal positive charges, look for the point where the field is zero. Why is it in the middle?
Show the contributions and put P at any point. Check that the orange arrow is the vector sum of the two dashed ones.

Charge Q₁

Charge Q₂

Field at P

Field of Q₁

Field of Q₂

Distance from P to Q₁

Force on a 1 nC charge

Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A 2 nC test charge, placed at a point, feels an electric force of \(6 \cdot 10^{-4}\ \text{N}\). What is the size of the electric field at that point?

Show solution
The field is the force per unit charge, \(E = \dfrac{F}{q} = \dfrac{6 \cdot 10^{-4}}{2 \cdot 10^{-9}}\).
\(E = 3 \cdot 10^5\ \text{N/C}\)
basic
What is the size of the electric field created by a point charge of \(4\ \mu\text{C}\) at a point 20 cm from it? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
For a point charge, \(E = k\,\dfrac{|Q|}{d^2}\) \(= 9 \cdot 10^9 \cdot \dfrac{4 \cdot 10^{-6}}{0.2^2}\).
\(E = 9 \cdot 10^5\ \text{N/C}\)
basic
At a point P, the electric field points to the right. In which direction is the force on a proton placed at P? And on an electron?

Show solution
The field is defined by the force on a positive charge, and the proton feels a force in the direction of the field.
A negative charge feels a force in the opposite direction.
To the right on the proton and to the left on the electron.
basic
At 10 cm from a point charge, the field is \(8 \cdot 10^4\ \text{N/C}\). What is it at 20 cm? And at 40 cm?

Show solution
The field of a point charge falls with the square of the distance. At double the distance, it is divided by 4; at four times the distance, by 16.
\(2 \cdot 10^4\ \text{N/C}\) at 20 cm and \(5 \cdot 10^3\ \text{N/C}\) at 40 cm
intermediate
An electron is in a region with an electric field of \(1000\ \text{N/C}\). What is the force on it and the acceleration that this force produces? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\) and \(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\).

Show solution
The force is \(F = e\,E = 1.6 \cdot 10^{-19} \cdot 1000\), in the direction opposite to the field.
By Newton's second law, \(a = \dfrac{F}{m} = \dfrac{e\,E}{m_e}\).
\(F = 1.6 \cdot 10^{-16}\ \text{N}\) and \(a \approx 1.8 \cdot 10^{14}\ \text{m/s}^2\)
intermediate
A charge of \(+2\ \mu\text{C}\) is at \(x = 0\) and another of \(-2\ \mu\text{C}\) at \(x = 20\ \text{cm}\). What is the electric field at the midpoint between them? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
Each charge is 10 cm from the midpoint and creates a field there of \(9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-6}}{0.1^2}\) \(= 1.8 \cdot 10^6\ \text{N/C}\).
The field of the positive charge points away from it, to the right, and that of the negative one points towards it, also to the right, and the two add up.
\(E = 3.6 \cdot 10^6\ \text{N/C}\), pointing towards the negative charge
intermediate
Two equal positive charges are a certain distance apart. At what point on the line joining them is the electric field zero? Why do two field lines never cross?

Show solution
At the midpoint, the two fields have the same size and opposite directions, and they cancel.
If two lines crossed, the field would have two directions at the crossing point, and a charge placed there would feel two different forces at the same time.
The field is zero at the midpoint, and the lines do not cross because the field has only one direction at each point.
intermediate
Two charges of \(+1\ \mu\text{C}\) are 30 cm from a point P, one in the horizontal direction and the other in the vertical, forming a right angle at P. What is the size of the field at P? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
Each charge creates at P a field of \(9 \cdot 10^9 \cdot \dfrac{10^{-6}}{0.3^2} = 1 \cdot 10^5\ \text{N/C}\).
The two vectors are perpendicular, and the size of the sum comes from Pythagoras' theorem, \(E = \sqrt{(10^5)^2 + (10^5)^2}\).
\(E \approx 1.4 \cdot 10^5\ \text{N/C}\), at 45° to each direction
challenge
A charge of \(+9\ \mu\text{C}\) is at \(x = 0\) and another of \(+1\ \mu\text{C}\) at \(x = 40\ \text{cm}\). At what point between them is the electric field zero?

Show solution
At the point we are looking for, the sizes of the two fields are equal, \(\dfrac{k \cdot 9}{x^2} = \dfrac{k \cdot 1}{(0.4 - x)^2}\).
Taking the square root, \(\dfrac{3}{x} = \dfrac{1}{0.4 - x}\), and so \(1.2 - 3x = x\).
At \(x = 30\ \text{cm}\), 10 cm from the smaller charge.
challenge
In one version of Millikan's experiment, an oil drop of mass \(1.6 \cdot 10^{-15}\ \text{kg}\) hangs still in the air between two plates, in a vertical electric field of \(5 \cdot 10^4\ \text{N/C}\). What is the charge on the drop, and how many extra electrons does it have, if the field points downwards? Use \(g = 10\ \text{m/s}^2\) and \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
While the drop is still, the electric force balances the weight, \(q\,E = m\,g\), and so \(|q| = \dfrac{1.6 \cdot 10^{-15} \cdot 10}{5 \cdot 10^4}\).
With the field pointing down, an upward force requires a negative charge, and the number of extra electrons is \(|q|/e\).
The drop has \(-3.2 \cdot 10^{-19}\ \text{C}\), that is, 2 extra electrons.

STEP 5

How much energy does the field give a charge?

The electric field of charges at rest, like the gravitational field, is conservative. The work it does on a charge that goes from A to B depends only on the two points, and so we can assign to each point an electric potential energy, which for two point charges is \(E_p = kQq/d\).

Dividing this energy by the charge \(q\), we get the electric potential, \(V = kQ/d\), measured in volts. Potential is a scalar quantity, with a sign, and with several charges we simply add up the potentials of each one, with no vectors.

\(V = k\,\dfrac{Q}{d}\)\(E_p = k\,\dfrac{Q\,q}{d}\)\(\tau_{AB} = q\,(V_A - V_B)\)\(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\)We take the potential as zero very far from the charges. In the simulation, the sources have \(\pm 2\ \mu\text{C}\), the screen represents 80 cm in width and the thin lines are the equipotentials of ±20, 40, 80, 160 and 320 kV. The work is the sum of \(q\,\vec E \cdot \Delta\vec s\) over thousands of small stretches of the path.

The work done by the field between two points is \(\tau = q\,(V_A - V_B)\), and the difference \(V_A - V_B\) is the same potential difference we measure in circuits, which we write as U, where many British books use V. Positive charges left free tend to move to lower potentials, and negative ones to higher potentials.

Equipotential surfaces gather the points at the same potential and cross the field lines at right angles, and when we carry a charge along one of them the field does no work. In atomic physics, we often measure energy in electronvolts; one electronvolt is the energy an electron gains in crossing 1 V, \(1.6 \cdot 10^{-19}\ \text{J}\).

Let's discuss

Carry the charge from A to B along the straight path and note the work. Repeat along the arc and the zigzag. What seems to repeat itself?
Drag B onto the same thin line that A is on. What is the work now?
With the dipole, put A and B on the dark line in the middle. Both potentials are zero; is the field there zero too?
Change the sign of the charge being carried. What happens to the sign of the work, and what does that suggest about how the charge would move on its own?

Charge carried

Potential at A

Potential at B

V_A − V_B

Work along the path

q·(V_A − V_B)

Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
What is the electric potential 30 cm from a point charge of \(+2\ \mu\text{C}\), taking the potential as zero very far from it? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
The potential of a point charge is \(V = k\,\dfrac{Q}{d} = 9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-6}}{0.3}\).
\(V = 6 \cdot 10^4\ \text{V}\)
basic
A charge of \(3\ \mu\text{C}\) is carried by the field from a point A at a potential of 500 V to a point B at 200 V. What is the work done by the electric force?

Show solution
The work done by the field is \(\tau = q\,(V_A - V_B)\) \(= 3 \cdot 10^{-6} \cdot (500 - 200)\).
\(\tau = 9 \cdot 10^{-4}\ \text{J}\)
basic
A charge is carried from one point to another on the same equipotential surface. What is the work done by the electric force? And if the path is long and full of bends?

Show solution
The work done by the field depends only on the potential difference between the starting and finishing points, \(\tau = q\,(V_A - V_B)\).
On the same equipotential, \(V_A = V_B\).
The work is zero, whatever the path.
basic
What is the electric potential energy of a charge of \(1\ \mu\text{C}\) 10 cm from another of \(2\ \mu\text{C}\)? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
For two point charges, \(E_p = k\,\dfrac{Q\,q}{d}\) \(= 9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-6} \cdot 1 \cdot 10^{-6}}{0.1}\).
\(E_p = 0.18\ \text{J}\)
intermediate
An electron starts from rest and is accelerated through a potential difference of 2000 V. What kinetic energy does it gain, in electronvolts and in joules, and what speed does it reach? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\) and \(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\).

Show solution
Each volt crossed gives the electron 1 eV, and in joules the energy is \(E_c = 2000 \cdot 1.6 \cdot 10^{-19}\).
From the kinetic energy, \(v = \sqrt{\dfrac{2\,E_c}{m}}\) \(= \sqrt{\dfrac{6.4 \cdot 10^{-16}}{9.1 \cdot 10^{-31}}}\).
2000 eV, or \(3.2 \cdot 10^{-16}\ \text{J}\), and \(v \approx 2.7 \cdot 10^7\ \text{m/s}\)
intermediate
Charges of \(+3\ \mu\text{C}\) and \(-3\ \mu\text{C}\) are 20 cm apart. At the midpoint, what is the potential? And the field? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
The potential is a scalar and adds up the contributions with their signs, \(V = \dfrac{k \cdot 3 \cdot 10^{-6}}{0.1}\) \(+ \dfrac{k \cdot (-3 \cdot 10^{-6})}{0.1}\).
The field is a vector, and both fields point towards the negative charge, \(E = 2 \cdot 9 \cdot 10^9 \cdot \dfrac{3 \cdot 10^{-6}}{0.1^2}\).
\(V = 0\) and \(E = 5.4 \cdot 10^6\ \text{N/C}\); zero potential at a point does not imply zero field.
intermediate
Two charges of \(+2\ \mu\text{C}\) are 40 cm apart. Work out the potential and the field at the midpoint. Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
Each charge, at 20 cm, contributes \(9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-6}}{0.2} = 9 \cdot 10^4\ \text{V}\), and the potentials add up.
The fields have the same size and opposite directions and cancel.
\(V = 1.8 \cdot 10^5\ \text{V}\) and \(E = 0\); zero field does not imply zero potential either.
intermediate
A charge of \(-2\ \mu\text{C}\) goes from a point A, at a potential of 100 V, to a point B, at 400 V. What is the work done by the electric force? Can this motion happen on its own?

Show solution
The work is \(\tau = q\,(V_A - V_B)\) \(= -2 \cdot 10^{-6} \cdot (100 - 400)\).
The work done by the field is positive, and the potential energy of the charge decreases, as with a falling object.
\(\tau = +6 \cdot 10^{-4}\ \text{J}\); negative charges do indeed tend to move to higher potentials.
challenge
How much work do we need to do to bring a charge of \(1\ \mu\text{C}\) slowly from very far away to 10 cm from a fixed charge of \(+5\ \mu\text{C}\)? What is the work done by the electric force along the way? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
Very far away, the potential energy is zero; at 10 cm, it is \(E_p = 9 \cdot 10^9 \cdot \dfrac{5 \cdot 10^{-6} \cdot 1 \cdot 10^{-6}}{0.1}\).
Bringing the charge slowly, with no gain in kinetic energy, our work equals the increase in potential energy, and the work of the electric force, which is repulsive, has the same size and the opposite sign.
We need to do \(+0.45\ \text{J}\), and the electric force does \(-0.45\ \text{J}\).
challenge
A fixed charge of \(+4\ \mu\text{C}\) creates a field around it. A 1 g bead, with a charge of \(+2\ \mu\text{C}\), is released from rest 20 cm from the fixed charge and moves away, pushed by the field. What is its speed as it passes 50 cm from the fixed charge? Ignore the weight and air resistance. Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
The potentials are \(V_A = 9 \cdot 10^9 \cdot \dfrac{4 \cdot 10^{-6}}{0.2}\) \(= 1.8 \cdot 10^5\ \text{V}\) and \(V_B = 9 \cdot 10^9 \cdot \dfrac{4 \cdot 10^{-6}}{0.5}\) \(= 7.2 \cdot 10^4\ \text{V}\).
The work done by the field, which becomes kinetic energy, is \(\tau = q\,(V_A - V_B)\) \(= 2 \cdot 10^{-6} \cdot (1.8 \cdot 10^5 - 7.2 \cdot 10^4)\) \(= 0.216\ \text{J}\), and \(v = \sqrt{\dfrac{2 \cdot 0.216}{10^{-3}}}\).
\(v \approx 21\ \text{m/s}\), whatever path the bead follows.

STEP 6

What happens between two charged plates?

Between two large, closely spaced parallel plates with opposite charges, the field is approximately uniform, with the same size, the same direction and the same sense at every point, from the positive plate to the negative one. In this case, the potential difference between the plates and the field are related by \(U = E\,d\).

When we launch a charged particle between the plates, it feels a constant force, \(F = qE\), perpendicular to them. If we ignore the weight, which for electrons and protons is very much smaller than the electric force, the motion repeats the horizontal projectile from Kinematics, with constant velocity along the plates and constant acceleration in the direction of the field, and the path is a parabola.

\(E = \dfrac{U}{d}\)\(F = q\,E\)\(a = \dfrac{q\,E}{m}\)\(t = \dfrac{L}{v_0}\)\(y = \dfrac{a\,t^2}{2}\)We use \(e = 1.6 \cdot 10^{-19}\ \text{C}\), \(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\) and \(m_p = 1.67 \cdot 10^{-27}\ \text{kg}\). In the simulation, the plates have \(L = 10\ \text{cm}\), the field is assumed uniform between them and zero outside, and the launch speed comes in multiples of \(10^7\ \text{m/s}\) for the electron and \(10^5\ \text{m/s}\) for the proton.

In a conductor in equilibrium, the excess charges sit on the surface and the field inside the metal is zero. For this reason, a car or an aeroplane struck by lightning protects the people inside, and we call the effect electrostatic shielding, or a Faraday cage.

On the surface, the charges build up more in pointed regions, where the field becomes strong enough to ionise the air. This is the action of points, which the lightning conductor makes use of to offer the discharge a safe path to the Earth, through a thick, well-earthed cable.

Let's discuss

Launch the electron and then the proton, with the same controls. Which plate does each one go towards, and why does the proton, almost 2000 times heavier, need to be launched much more slowly for the deflection to show?
Double the potential difference and watch the deflection. Then, with the potential difference fixed, double the distance between the plates. What happens to the field?
Increase the launch speed. The deflection seems to drop a lot; why, if the force stays the same?
With \(v_0 = 1 \cdot 10^7\ \text{m/s}\) and \(d = 5\ \text{cm}\), find the smallest potential difference at which the electron hits the plate and check the value with \(y = a\,t^2/2\).

Potential difference

Distance between the plates

Initial speed (m/s)

Field

Force

Acceleration

Time between the plates

Deflection on exit

Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Two parallel plates, 2 cm apart, are connected to a 100 V supply. What is the electric field between them, assuming it is uniform?

Show solution
In a uniform field, \(U = E\,d\), and so \(E = \dfrac{U}{d} = \dfrac{100}{0.02}\).
\(E = 5000\ \text{V/m}\), the same as 5000 N/C
basic
During a thunderstorm, why is it safer to be inside a closed car, with a metal body, than out in the open?

Show solution
The metal body of the car is a conductor. The charges brought by a lightning strike spread over the outer surface and drain to the ground, and the field inside a conductor in equilibrium is zero.
The car works as a Faraday cage and shields the people inside; the tyres play no part in this protection.
basic
Why is the tip of a lightning conductor sharp, and why does it need a thick wire connected to the Earth?

Show solution
On a charged conductor, the charges concentrate in pointed regions, where the field becomes very strong and ionises the air, creating a preferred path for the discharge, the action of points.
The thick wire, of low resistance, carries the enormous current of the lightning to the Earth without heating up too much.
The tip attracts the discharge because of the action of points, and the earthing carries the lightning safely to the ground.
basic
Between two parallel plates 3 mm apart, the field is uniform and equal to \(2 \cdot 10^4\ \text{V/m}\). What is the potential difference between the plates?

Show solution
In a uniform field, \(U = E\,d = 2 \cdot 10^4 \cdot 3 \cdot 10^{-3}\).
\(U = 60\ \text{V}\)
intermediate
An electron is between two plates where the field is uniform and equal to \(1 \cdot 10^4\ \text{V/m}\). Compare the electric force on it with its weight. Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\), \(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\) and \(g = 10\ \text{m/s}^2\).

Show solution
The electric force is \(F = e\,E\) \(= 1.6 \cdot 10^{-19} \cdot 10^4\) \(= 1.6 \cdot 10^{-15}\ \text{N}\), and the weight is \(P = m\,g = 9.1 \cdot 10^{-30}\ \text{N}\).
The ratio is \(\dfrac{1.6 \cdot 10^{-15}}{9.1 \cdot 10^{-30}}\).
The electric force is about \(1.8 \cdot 10^{14}\) times the weight, which we can ignore.
intermediate
An electron enters horizontally, at \(2 \cdot 10^7\ \text{m/s}\), between two plates 5 cm long, where the field is vertical and equal to 2000 V/m. How far is it deflected vertically by the time it leaves the plates? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\) and \(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\).

Show solution
Horizontally, the velocity is constant, and the time between the plates is \(t = \dfrac{0.05}{2 \cdot 10^7}\) \(= 2.5 \cdot 10^{-9}\ \text{s}\).
Vertically, \(a = \dfrac{e\,E}{m}\) \(= \dfrac{1.6 \cdot 10^{-19} \cdot 2000}{9.1 \cdot 10^{-31}}\) \(\approx 3.5 \cdot 10^{14}\ \text{m/s}^2\), and the deflection is \(y = \dfrac{a\,t^2}{2}\), as in horizontal projectile motion.
\(y \approx 1.1 \cdot 10^{-3}\ \text{m}\), about 1.1 mm
intermediate
A proton starts from rest next to the positive plate of a pair of parallel plates connected to 200 V. With what kinetic energy and what speed does it reach the negative plate? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\) and \(m_p = 1.67 \cdot 10^{-27}\ \text{kg}\).

Show solution
The work done by the field, \(\tau = q\,U = 1.6 \cdot 10^{-19} \cdot 200\), becomes kinetic energy.
The speed comes from \(v = \sqrt{\dfrac{2\,E_c}{m}}\) \(= \sqrt{\dfrac{6.4 \cdot 10^{-17}}{1.67 \cdot 10^{-27}}}\).
\(E_c = 3.2 \cdot 10^{-17}\ \text{J}\) and \(v \approx 2.0 \cdot 10^5\ \text{m/s}\)
intermediate
Dry air stops being an insulator, and a spark jumps, when the field exceeds about \(3 \cdot 10^6\ \text{V/m}\), an approximate value. What is the largest potential difference we can apply between two flat electrodes 1 cm apart, in dry air, without a spark?

Show solution
Assuming the field is uniform between the electrodes, \(U = E\,d = 3 \cdot 10^6 \cdot 0.01\).
About \(3 \cdot 10^4\ \text{V}\), or 30 kV; with pointed electrodes, the spark jumps at lower potential differences.
challenge
An electron enters horizontally, at \(1 \cdot 10^7\ \text{m/s}\), between plates 10 cm long, in a vertical field of 2000 V/m. Work out the vertical deflection on exit and the angle the velocity makes with the horizontal at that instant. Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\) and \(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\).

Show solution
The time between the plates is \(t = \dfrac{0.1}{10^7} = 10^{-8}\ \text{s}\), and the acceleration is \(a = \dfrac{1.6 \cdot 10^{-19} \cdot 2000}{9.1 \cdot 10^{-31}}\) \(\approx 3.5 \cdot 10^{14}\ \text{m/s}^2\).
The deflection is \(y = \dfrac{a\,t^2}{2}\) \(\approx 1.8 \cdot 10^{-2}\ \text{m}\), and the vertical velocity on exit is \(v_y = a\,t\) \(\approx 3.5 \cdot 10^6\ \text{m/s}\).
The angle comes from \(\tan\theta = \dfrac{v_y}{v_0} \approx 0.35\).
A deflection of about 1.8 cm and an angle of about 19°
challenge
A hollow metal sphere of radius 10 cm has a charge of \(+2\ \text{nC}\), in equilibrium. What is the electric field inside it, just outside the surface and 20 cm from the centre? What is the potential at the centre? Use \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\).

Show solution
Inside a conductor in equilibrium, the field is zero. Outside, the sphere behaves like a point charge at the centre, and just outside the surface \(E = 9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-9}}{0.1^2}\).
At 20 cm, the field is \(9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-9}}{0.2^2}\), a quarter as much. Since the field inside is zero, the potential is the same throughout the interior, equal to that of the surface, \(V = 9 \cdot 10^9 \cdot \dfrac{2 \cdot 10^{-9}}{0.1}\).
Zero field inside, 1800 N/C just outside and 450 N/C at 20 cm; a potential of 180 V at the centre.

STEP 7

How can we store energy in an electric field?

A capacitor is made of two conductors close together, separated by an insulator. When we connect it to a battery, it receives a charge \(+Q\) on one plate and \(-Q\) on the other, proportional to the potential difference. The constant of proportionality is the capacitance, \(C = Q/U\), measured in farads.

To charge the capacitor, the battery does work against the repulsion of the charges already on the plates, and this work is stored in the field between them. The energy is \(QU/2\), or \(CU^2/2\), and the half appears because the potential difference grows from zero to \(U\) as the charges arrive.

In a parallel-plate capacitor, the capacitance grows with the area of the plates and decreases with the distance between them, \(C = \varepsilon A/d\), where \(\varepsilon\) depends on the insulator. One possible intuition is that larger plates hold more charge at the same potential difference, and plates closer together let the attraction between the opposite charges hold on to more charge.

A capacitor gathers energy slowly and gives it back all at once. In a camera flash, a capacitor of typically a hundred to a few hundred microfarads, charged to about 300 V, stores a few joules, which the flash tube uses up in about a thousandth of a second. A defibrillator does something similar on a larger scale, with a few hundred joules delivered in a few milliseconds, probably the most dramatic use of a capacitor.

\(C = \dfrac{Q}{U}\)\(E = \dfrac{Q\,U}{2} = \dfrac{C\,U^2}{2}\)\(C = \varepsilon\,\dfrac{A}{d}\)The values for the flash and the defibrillator are approximate and vary from one device to another. In the simulation, the capacitor has \(100\ \mu\text{F}\) with the area and the distance at ×1, and the capacitance follows \(C \propto A/d\). The battery charges it through a 5 kΩ resistor, and the discharge through the lamp is shown in slow motion.

Let's discuss

Charge the capacitor to 300 V and check \(Q = C\,U\) and \(E = C\,U^2/2\) in the readings. Then discharge it through the lamp.
With the battery connected, double the area of the plates. What happens to the charge and to the energy?
Charge it, open the switch and only then move the plates apart. Why is it now the potential difference that rises, rather than the charge that changes?
Compare the energy at 150 V and at 300 V. Why does double the potential difference give four times the energy?

Battery p.d.

Area of the plates

Distance between the plates

Capacitance

P.d. across the capacitor

Charge

Energy stored

Energy of the last flash

Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A \(500\ \mu\text{F}\) capacitor is connected to a 12 V battery. What charge does it store?

Show solution
From the definition of capacitance, \(Q = C\,U = 500 \cdot 10^{-6} \cdot 12\).
\(Q = 6 \cdot 10^{-3}\ \text{C}\), or 6 mC
basic
The capacitor in a camera flash has \(100\ \mu\text{F}\) and is charged to 300 V. How much energy does it store?

Show solution
The energy of a capacitor is \(E = \dfrac{C\,U^2}{2} = \dfrac{100 \cdot 10^{-6} \cdot 300^2}{2}\).
\(E = 4.5\ \text{J}\)
basic
In a parallel-plate capacitor, what happens to the capacitance if we double the area of the plates? And if we double the distance between them?

Show solution
The capacitance of parallel plates is \(C = \varepsilon\,\dfrac{A}{d}\), proportional to the area and inversely proportional to the distance.
With double the area, the capacitance doubles; with double the distance, it halves.
basic
A capacitor connected to 50 V stores a charge of \(2 \cdot 10^{-4}\ \text{C}\). What is its capacitance?

Show solution
The capacitance is \(C = \dfrac{Q}{U} = \dfrac{2 \cdot 10^{-4}}{50}\).
\(C = 4 \cdot 10^{-6}\ \text{F}\), or \(4\ \mu\text{F}\)
intermediate
A defibrillator uses a capacitor of about \(32\ \mu\text{F}\) charged to some 5000 V, approximate values that vary with the model. How much energy does it store, and what is the charge?

Show solution
The energy is \(E = \dfrac{C\,U^2}{2} = \dfrac{32 \cdot 10^{-6} \cdot 5000^2}{2}\), and the charge is \(Q = C\,U = 32 \cdot 10^{-6} \cdot 5000\).
About 400 J, with a charge of 0.16 C.
intermediate
A capacitor connected to a battery goes from 6 V to 12 V when we change the battery. What happens to the charge and the energy stored?

Show solution
The charge \(Q = C\,U\) is proportional to the potential difference and doubles.
The energy \(C\,U^2/2\) is proportional to the square of the potential difference and is multiplied by \(2^2 = 4\).
The charge doubles and the energy becomes four times as large.
intermediate
The capacitor of a flash stores 4.5 J and delivers it to the flash tube in about 1 ms. What is the average power of the flash? Compare it with a 9 W LED lamp.

Show solution
The average power is the energy divided by the time, \(P = \dfrac{4.5}{10^{-3}}\).
The ratio to the LED lamp is \(\dfrac{4500}{9}\).
About 4500 W, some 500 times the power of the LED lamp, for a thousandth of a second.
intermediate
A \(10\ \mu\text{F}\) capacitor is charged to 100 V and disconnected from the battery. We then move the plates apart to double the distance. Work out the charge, the potential difference and the energy before and after.

Show solution
Before, \(Q = C\,U = 10^{-5} \cdot 100 = 10^{-3}\ \text{C}\) and \(E = \dfrac{10^{-5} \cdot 100^2}{2} = 0.05\ \text{J}\).
Disconnected, the capacitor keeps its charge. With double the distance, \(C\) drops to \(5\ \mu\text{F}\), and \(U = \dfrac{Q}{C} = \dfrac{10^{-3}}{5 \cdot 10^{-6}}\) and \(E = \dfrac{Q\,U}{2}\).
The charge does not change, the potential difference doubles to 200 V and the energy doubles to 0.1 J, thanks to the work done to move the plates apart.
challenge
A camera uses two 1.5 V AA cells in series, each with about 2500 mAh, to charge the flash capacitor, at 4.5 J per flash. If all the energy of the cells went to the flash, how many flashes would be possible? Why, in practice, is the number much smaller?

Show solution
Each cell stores \(E = q\,U\) \(= 2.5 \cdot 3600 \cdot 1.5\) \(= 13\,500\ \text{J}\), and the two together 27,000 J.
The number of flashes would be \(\dfrac{27\,000}{4.5}\).
About 6000 flashes in the ideal case; in practice, the circuit that raises the potential difference to 300 V loses energy, the camera uses the cells for other functions and the cell does not deliver all its charge.
challenge
A parallel-plate capacitor in air has plates of \(1\ \text{m}^2\) separated by 1 mm. What is its capacitance? What area would be needed, with the same distance, to reach 1 F? Use \(\varepsilon_0 = 8.85 \cdot 10^{-12}\ \text{F/m}\).

Show solution
From the parallel-plate capacitor formula, \(C = \varepsilon_0\,\dfrac{A}{d}\) \(= 8.85 \cdot 10^{-12} \cdot \dfrac{1}{10^{-3}}\).
For 1 F, \(A = \dfrac{C\,d}{\varepsilon_0}\) \(= \dfrac{1 \cdot 10^{-3}}{8.85 \cdot 10^{-12}}\).
About 8.9 nF, and an area of \(1.1 \cdot 10^8\ \text{m}^2\) would be needed, more than 100 km²; the farad is an enormous unit.

WRAP-UP

Challenges

Electric charge

In a lightning strike, about 20 C of negative charge pass from the cloud to the ground. How many electrons is that? (\(e = 1.6 \cdot 10^{-19}\ \text{C}\))

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The number of electrons is the charge divided by the charge of each one, \(n = \dfrac{20}{1.6 \cdot 10^{-19}}\).
That gives \(1.25 \cdot 10^{20}\) electrons, more than a hundred billion billion.

Charging by contact

Two identical metal spheres, with +10 nC and −4 nC, touch and are separated. Then one of them touches a third identical, neutral sphere. What is the final charge on each sphere?

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In the first contact, each sphere ends up with \(\dfrac{10 - 4}{2}\), or 3 nC.
In the second, the 3 nC sphere shares its charge with the neutral one, and the three end up with 3, 1.5 and 1.5 nC, which add up to the initial 6 nC.

Coulomb's law

Two charges attract each other with 0.9 N when they are 10 cm apart. At what distance does the force drop to 0.1 N?

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The force became 9 times smaller, and since it falls with \(1/d^2\), the distance has to grow \(\sqrt{9}\) times.
The new distance is \(3 \cdot 10\), or 30 cm.

Electric field

Near the ground, in fair weather, the atmosphere has an electric field of about 100 N/C, an approximate value. Compare the electric force on an electron with its weight. (\(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\), \(g = 10\ \text{m/s}^2\))

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The electric force is \(F = e\,E\) \(= 1.6 \cdot 10^{-19} \cdot 100\) \(= 1.6 \cdot 10^{-17}\ \text{N}\), and the weight is \(m\,g = 9.1 \cdot 10^{-30}\ \text{N}\).
The ratio, \(\dfrac{1.6 \cdot 10^{-17}}{9.1 \cdot 10^{-30}}\), comes to about \(1.8 \cdot 10^{12}\); even a weak field dominates the motion of the electron.

Electric potential

At what distance from a point charge of 1 µC is the potential 9000 V? What shape is the 9000 V equipotential surface? (\(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\))

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Isolating the distance in \(V = kQ/d\), we have \(d = \dfrac{9 \cdot 10^9 \cdot 10^{-6}}{9000}\), or 1 m.
Every point 1 m from the charge has this potential, and the equipotential is a sphere of radius 1 m, centred on the charge.

Uniform field

During a thunderstorm, the base of a cloud 1 km above the ground may be at about \(10^8\) V relative to the ground. If we treat the cloud and the ground as parallel plates, what, roughly, is the average field? Compare it with the \(3 \cdot 10^6\ \text{V/m}\) at which dry air stops insulating.

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With \(E = U/d\), the average field is \(\dfrac{10^8}{1000} = 10^5\ \text{V/m}\), some 30 times smaller than the limit for air.
Lightning starts where the local field becomes much larger than the average, near droplets, ice crystals and points, and the plate model only lets us estimate the order of magnitude of the field.

Capacitors

A 1 F supercapacitor is charged to 2.7 V. How much energy does it store? Compare it with the 4.5 J of a flash capacitor, of 100 µF at 300 V.

Show solution

The energy is \(E = \dfrac{C\,U^2}{2} = \dfrac{1 \cdot 2.7^2}{2}\), about 3.6 J.
The capacitance is ten thousand times larger, and even so the energy falls below that of the flash, because the energy grows with the square of the potential difference.

Thinking, no sums

On a dry day, Marcela gets out of the car and feels a small shock when she touches the door. Why does this happen, and why is it rarer on humid days?

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As she slides across the seat, her body and clothes exchange electrons with the fabric and become charged by friction.
When she touches the metal of the door, the built-up charge drains away all at once, in a small spark. On humid days, a thin layer of water on the surfaces conducts a little, and the charge tends to leak away gradually, before it builds up.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, and we wrote the five questions below in its format, with an everyday context, a question and five options, each one returning to a different step of the lesson. Further down we list real exam questions on electrostatics, a topic that tends to appear in connection with lightning, shocks and household appliances.

Charging by friction · Step 2

In a science lesson, Thiago rubbed a PVC drinking straw with a woollen cloth and then rubbed a rubber balloon on his own hair. The teacher showed the class the part of a triboelectric series reproduced below, reminding them that the order of the materials may vary a little from one source to another.

From the material that gives up electrons most easily to the one that takes them most easily
Position	Material
1	dry human skin
2	glass
3	human hair
4	wool
5	silk
6	cotton
7	rubber
8	PVC

Based on the series, which statement correctly describes what happened?

The straw became positive, because it received protons from the wool.
The straw became negative and the wool positive, because electrons passed from the wool to the straw.
The straw and the wool both became negative, because the friction created electrons in both.
The balloon became positive and the hair negative, because hair comes before rubber in the series.
The wool became negative, because it is closer to the start of the series than PVC.

Show solution

Answer: B.
When two materials are rubbed together, the one that comes earlier in the series gives up electrons to the one that comes later. The wool (position 4) gives up electrons to the PVC (position 8), and the straw becomes negative, while the wool becomes positive, with a charge of the same size.
Option A talks of protons, which stay locked in the nuclei, and C assumes that friction creates charge, which goes against conservation. Option D reverses the result, since hair gives up electrons to rubber and the balloon becomes negative, and E reads the series backwards.
The same reasoning suggests that pairs of materials further apart in the series tend to exchange more charge, although the humidity of the air and how hard they are rubbed also count for a lot.

Coulomb's law · Step 3
In 1785, Coulomb measured the force between small charged spheres with a torsion balance. In a reconstruction of the experiment, Larissa measures a repulsive force of \(4.0 \cdot 10^{-4}\ \text{N}\) between two small metal spheres 6 cm apart. She then touches one of the spheres with a third, identical and neutral, and takes it away, which halves the charge on the first. Finally, she brings the two original spheres closer, to 3 cm. Let us assume they behave like point charges.

What is the new force between the spheres?
1. \(1.0 \cdot 10^{-4}\ \text{N}\)
2. \(2.0 \cdot 10^{-4}\ \text{N}\)
3. \(4.0 \cdot 10^{-4}\ \text{N}\)
4. \(8.0 \cdot 10^{-4}\ \text{N}\)
5. \(1.6 \cdot 10^{-3}\ \text{N}\)
Show solution

Answer: D.
By Coulomb's law, \(F = k\,|q_1\,q_2|/d^2\). Halving the charge divides the force by 2, and halving the distance multiplies it by \(2^2 = 4\).
The new force is \(F^{\prime} = 4.0 \cdot 10^{-4} \cdot \dfrac{4}{2}\), or \(8.0 \cdot 10^{-4}\ \text{N}\).
Option B takes only the charge into account, and E only the distance. Option C treats the force as inversely proportional to the distance, without the square, and A reverses the effect of bringing the spheres closer.
Potential and the electronvolt · Step 5
In radiotherapy, a linear accelerator produces beams of electrons to treat tumours. In one of its settings, each electron leaves the accelerator with an energy of 6 MeV, the same as it would gain by starting from rest and crossing a potential difference of 6 million volts. The elementary charge is \(1.6 \cdot 10^{-19}\ \text{C}\).

What is the energy of each electron in the beam, in joules?
1. \(2.7 \cdot 10^{-26}\ \text{J}\)
2. \(1.6 \cdot 10^{-19}\ \text{J}\)
3. \(9.6 \cdot 10^{-19}\ \text{J}\)
4. \(9.6 \cdot 10^{-13}\ \text{J}\)
5. \(3.75 \cdot 10^{25}\ \text{J}\)
Show solution

Answer: D.
The energy a charge gains in crossing a potential difference is \(E = q\,U\), and for the electron, \(E = 1.6 \cdot 10^{-19} \cdot 6 \cdot 10^6\).
That gives \(9.6 \cdot 10^{-13}\ \text{J}\), a minute amount for us and an enormous one for a single electron.
Option C forgets the mega prefix and works out 6 eV, and B corresponds to 1 eV. Option A divides the charge by the potential difference, and E divides the potential difference by the charge.
Electrostatic shielding · Step 6
When maintaining power lines without switching off the supply, some electricians reach the cable in an insulated basket mounted on a lorry and wear overalls woven with metal threads, connected to the cable itself. They touch cables at hundreds of thousands of volts and get no shock.

What is the physical explanation for the electrician's protection?
1. The overalls insulate the body and stop current from flowing.
2. The conducting overalls are at the same potential as the cable, and the electric field inside them is practically zero.
3. The potential difference of the cable drops to zero the moment the electrician touches it.
4. The metal threads of the overalls work as a lightning conductor and carry the charge of the cable to the Earth.
5. The current passes through the electrician's body, but it is harmless because it is alternating.
Show solution

Answer: B.
The overalls are a conductor that surrounds the body, like a Faraday cage. Connected to the cable, they are at the same potential as it, and inside a conductor in equilibrium the electric field is zero, so that there is no potential difference between parts of the body.
The insulated basket blocks any path to the Earth, and that is why D fails. Option A mistakes the overalls for insulating clothing, C imagines that the potential difference of the cable vanishes, and E is dangerous, because alternating current through the body can kill.
Capacitors · Step 7
Caio took apart the emergency light of a bicycle and found a capacitor that stores energy for a few seconds of light. He measured the charge on the capacitor for several potential differences and drew the graph below.

How much energy does the capacitor store when it is charged to 12 V?
1. \(1.0 \cdot 10^{-4}\ \text{J}\)
2. \(6.0 \cdot 10^{-4}\ \text{J}\)
3. \(7.2 \cdot 10^{-3}\ \text{J}\)
4. \(1.44 \cdot 10^{-2}\ \text{J}\)
5. \(7.2\ \text{J}\)
Show solution

Answer: C.
The energy stored is the area under the line of the \(Q \times U\) graph, a triangle, \(E = \dfrac{Q\,U}{2}\).
With 12 V and \(1.2\ \text{mC} = 1.2 \cdot 10^{-3}\ \text{C}\), \(E = \dfrac{1.2 \cdot 10^{-3} \cdot 12}{2}\), or \(7.2 \cdot 10^{-3}\ \text{J}\).
Option D forgets the half, and E forgets the milli prefix. Option A is the value of the capacitance, \(C = Q/U = 10^{-4}\ \text{F}\), and B divides the charge by two without multiplying by the potential difference.
A few millijoules may seem little, but they are enough to keep an LED lit for a few seconds.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where we can look each question up by year, day, booklet colour and number.

ENEM 2010, Day 1, blue booklet, question 78. Two sisters keep their mobile phones in boxes of different materials, one metal and the other wood, and only one of the phones receives calls; the question asks for the material of the box that blocked the signal and the explanation, electrostatic shielding.
ENEM 2013, Day 1, blue booklet, question 79. The lamp in a circuit with a cell lights up almost the instant we close the switch, and we need to say what explains this speed in the classical model of current, the electric field that sets up throughout the circuit.
ENEM 2020, Day 2, blue booklet, question 91. A comic strip shows a cat that rubs itself against a pair of trousers and becomes charged, and we choose the mechanism of charging by friction, the transfer of electrons from one material to the other.
ENEM 2020, Day 2, blue booklet, question 133. The text recommends the inside of a closed car as a shelter during thunderstorms with lightning, and the question asks for the physical reason for this protection, the shielding by the metal body.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Quantised charge\(Q = n\,e\)

Constants\(e = 1.6 \cdot 10^{-19}\ \text{C}\), \(k = 9 \cdot 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2\)

Contact (identical spheres)\(Q^{\prime} = (Q_A + Q_B)/2\)

Coulomb's law\(F = k\,|q_1 q_2|/d^2\)

Electric field\(\vec E = \vec F/q\)

Field of a point charge\(E = k\,|Q|/d^2\)

Potential\(V = k\,Q/d\), \(E_p = k\,Q\,q/d\)

Work done by the field\(\tau = q\,(V_A - V_B)\)

Uniform field\(U = E\,d\)

Capacitance\(C = Q/U\), \(C = \varepsilon A/d\)

Energy in a capacitor\(E = Q\,U/2 = C\,U^2/2\)

Electronvolt\(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\)