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18.1Electric charge, force and field
Electric charge is a property of matter. Ordinarily, matter appears electrically neutral, but we can charge a body by friction.
For example, take two plastic rods and rub each with a piece of wool. We will find that the two rods now repel each other. If we now rub two glass rods with silk, we find that the glass rods again repel each other, but the charged glass rod attracts the charged plastic rod. We can understand these observations (Figure 18.1) by assuming that:
âthe process of rubbing involves the transfer of charge from one body to the other
âcharge can be positive or negative
âthere is a force between charged bodies that can be attractive or repulsive.

Figure 18.1: Two simple experiments to investigate properties of electric charge.

Benjamin Franklin (1706â1790) decided to call the sign of the charge on the glass rubbed with silk âpositiveâ. Much later, when electrons were discovered, it was found that electrons were attracted to the charged glass rod. This means that electrons must have negative charge. But if Franklin had called the charge on the glass rod negative, we would now be calling the electronâs charge positive!
From experiments with charged objects, we learn that there is a force of attraction between charges of opposite sign and a force of repulsion between charges of the same sign. The magnitude of the force becomes smaller as the distance between the charged bodies increases.
Properties of electric charge
In ordinary matter, negative charge is a property of particles called electrons. Positive charge is a property of protons, which exist in the nuclei of atoms. (There are many other particles that have charge, but those do not appear in ordinary matter.)
The first important property of electric charge is that it is conserved. Like total energy, electric charge cannot be created or destroyed. In any process the total charge cannot change (see worked example 18.1).
In solid metals the atoms are fixed in position in a lattice, but there are many âfreeâ electrons that do not belong to a particular atom. When exposed to an electric field (see later) these electrons can drift in the same direction creating an electric current. In liquids, and especially in gases, positive ions can also transport charge.
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Materials that have many âfreeâ electrons (Figure 18.2) are called conductors.
Materials that very few âfreeâ electrons, so charge cannot move freely, are called insulators.

Figure 18.2: In a conductor there are many âfreeâ electrons that move around much like molecules of a gas.

WORKED EXAMPLE 18.1
Two separated, identical conducting spheres are charged with charges of 4.0 ÂµC and â12 ÂµC, respectively. The spheres are allowed to touch and then are separated again. Determine the charge on each sphere. Describe the transfer of charge from one sphere to the other.
Answer
The net charge on the two spheres is 4.0 â 12 = â8.0 ÂµC. The contact of the two conducting spheres implies that charge will be transferred from one to the other. By symmetry, when the spheres are allowed to touch, they will end up with the same charge, since they are identical. The total amount of charge on the two spheres after separation must be â8.0 ÂµC by charge conservation.
When they separate, each will therefore have a charge of â4.0 ÂµC.
It is negative charge that gets transferred (electrons). The positive charges have fixed positions and do not move. So an amount of â8.0 ÂµC gets transferred.
The second important property of electric charge is that it is a quantised quantity; this means that the amount of electric charge on a body is always an integral multiple of a basic unit. The basic unit is the magnitude of the charge on the proton.
This amount of charge is symbolised by e. The charge on an electron is âe. (If we include quarks, the particles inside protons and neutrons, then the basic unit of charge is e3.) The SI unit of charge is the coulomb (C). 1 e = 1.6 Ã 10â19 C.
The quantisation of electric charge was determined in an experiment by Robert Millikan and is described at the end of this chapter.
Electrostatic induction
Friction is one way to charge a body. Electrostatic induction is another. Consider a positively charged rod that is placed close to, but not touching, a conducting sphere on an insulating stand.

Figure 18.3: Charging a sphere by induction.

The positive charge on the rod will attract free electrons in the sphere closer to the rod. This means the left side of the sphere will be left with an excess of positive charge. The net charge on the sphere is still zero. In the second diagram the sphere is grounded. This means that we connect the sphere to the ground with a cable. Electrons from the earth will move through the cable and neutralize some of the positive charge on the left side of the sphere. The connection to the earth and the charged rod are both removed. The sphere is left with a net negative charge that will distribute itself uniformly on the surface of the sphere. The sphere is left with a net charge that is opposite to that on the charged rod.
If the rod is negatively charged, the negative charge will push electrons away leaving the right side of the sphere with an excess of positive charge. Grounding (earthing) will make electrons move to the earth leaving the sphere with a net positive charge after grounding and rod are removed. Again the charge on the sphere is opposite to that on the rod.
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CHECK YOURSELF 1
A student says, referring to Figure 18.3, that when the positively charged rod is put close to the sphere, negative charge is attracted to the right side of the sphere and positive charge is repelled to the left side. Comment on this statement.
Coulombâs law for the electric force
The electric force between two electric charges, q1 and q2, was investigated in 1785 by Charles Augustin Coulomb (1736â1806). Coulomb discovered that this force is inversely proportional to the square of the separation of the charges and is proportional to the product of the two charges. It is attractive for charges of opposite sign and repulsive for charges of the same sign (Figure 18.4).

Figure 18.4: The force between two point electric charges is given by Coulombâs law and can be attractive or repulsive.

Coulombâs law states that the electric force F between two point charges q1 and q2 separated by a distance r is given by:
F=kq1q2r2
The constant k is known as the Coulomb constant and equals 8.99 Ã 109 NCâ2 m2 in vacuum.
The constant k is also written as k=14ÏÎµ where Îµ is known as permittivity. If the charges are surrounded by vacuum, the value of the permittivity is Îµ0 = 8.85 Ã 10â12 C2 Nâ1 mâ2. If they are surrounded by some other medium, we must use the value of the permittivity appropriate to that medium.
WORKED EXAMPLE 18.2
Two charges, q1 = 2.0 ÂµC and q2 = 8.0 ÂµC, are placed along a straight line separated by a distance of 3.0 cm.
aCalculate the force exerted on each charge in vacuum.
bCalculate the force when the charges are surrounded by water whose permittivity is 80 times that of the vacuum.
cThe charge q1 is increased to 4.0 ÂµC. Determine the force on each charge now (in vacuum).
Answer
aThis is a straightforward application of the formula F=kq1q2r2. We find that:
F=8.99Ã109Ã2.0Ã8.0Ã10â129.0Ã10â4
F=160Â N
This is the force that q1 exerts on q2, and vice versa.
bThe Coulomb constant is now 80 smaller so the force is 2.0 N.
cSince the charge doubles the force doubles to F = 320 N on both charges.
EXAM TIP
It is a common mistake to double the force on one charge but not the other.
WORKED EXAMPLE 18.3
A positive charge q is placed on the line joining q1 and q2 from worked example 2.2. Determine the distance from q1 where this third positive charge experiences zero net force.
Answer
Let that distance be x. A positive charge q at that point would experience a force from q1 equal to F1=kq1qx2 and a force in the opposite direction from q2 equal to F2=kq2q(dâx)2 where d = 3.0 cm is the distance between q1 and q2.
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Figure 18.5: For worked example 18.3.

Charge q will experience no net force when F1 = F2, so:
kq1qx2=kq2q(dâx)2
Cancelling kq and substituting q1 = 2.0 ÂµC and q2 = 8.0 ÂµC gives:
2.0x2=8.0(dâx)2
(dâx)2=4x2
(dâx)=2x
x=d3=1.0Â cm
EXAM TIP
We do not have to change units to C. The units on both sides of the equation are the same (ÂµC) and cancel out.
Electric field
The space around a charge or an arrangement of charges is different from space in which no charges are present. It contains an electric field. We can test whether an electric field exists at a point P by bringing a small, point charge q at P. If q experiences an electric force, then there is an electric field at P. If no force is experienced, then the electric field is zero. For this reason the small charge is called a test charge: it tests for the existence of electric fields. It has to be small so that its presence does not disturb the electric field it is trying to detect. By convention test charges are assumed to be positive.
KEY POINT
We define electric field strength, E, as the electric force per unit charge experienced by a small, positive, point charge q:
EâÂ =FâÂ q
Note that electric field strength is a vector quantity. The direction of the electric field is the same as the direction of the force experienced by a positive charge at the given point. The unit of electric field is N Câ1.
The magnitude of the force experienced by a test charge q placed a distance r from a point charge Q is (by Coulombâs law):
F=kQqr2
and so from the definition EâÂ =FâÂ q, the magnitude of the electric field strength is:
E=k(Qqr2)q
E=kQr2
The formula E=kQr2 applies to a point charge Q but also outside of a spherical charge. If Q is a spherical charge, r is the distance from the centre of the spherical charge.
CHECK YOURSELF 2
The distance of a point from a positive charge is doubled. What happens to the electric field strength?
The left diagram in Figure 18.6a shows the electric field at various points around a positive point charge. The left diagram in Figure 18.6b shows the electric field around a negative charge. The diagrams show the direction of the electric field strength, and the fact that the arrows get smaller as we move away from the charge indicates that the magnitude of the electric field strength decreases as demanded by the formula E=kQr2. Another way of representing the electric field is to imagine smooth lines going through the arrows. These are called electric field lines. This results in the diagrams on the right in Figure 18.6.
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Figure 18.6: The electric force experienced by a positive test charge at various positions near a a positive and b a negative point charge. The diagrams on the right show the electric field lines.

The disadvantage is that we lost the information on the magnitude of the electric field strength. We make up for this by arguing that the lines are further apart as we move away. So we take the density of the lines to be a measure of the magnitude of the electric field strength.
Figure 18.7a shows the field lines created by two equal, positive charges and Figure 18.7b by two equal and opposite charges.

Figure 18.7: Field lines for a two equal positive charges and b two equal and opposite charges.

Unlike the case of a single charge, the field lines are no longer straight. The electric field is tangential to the field lines in the direction of the arrow, Figures 18.8a and 18.8b. Thus the electric force is also tangential to the field lines. The force on a positive charge is in the direction of the electric field, and the force on a negative charge is opposite to the electric field. The magnitude of the force is F = qE.

Figure 18.8: a The electric force on a positive charge is tangent to the field line in the direction of the arrow of the field line. b The force on a negative charge is tangent to the field line in a direction opposite to the arrow. c The density of field lines increases as we move towards the top of the page and so the electric field strength increases.

CHECK YOURSELF 3
A negative charge is placed on the field line as shown in Figure 18.9.

Figure 18.9: For check yourself question 3.

Draw an arrow to show the electric force on the charge.
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Figure 18.10a shows the field lines created by two unequal, positive charges and Figure 18.10b the field created by two unequal and opposite charges. In Figure 18.10a q2 = 4q1 and in Figure 18.10b q2 = â4q1.

Figure 18.10: a Field lines for unequal and positive charges. b Field lines for unequal and opposite charges.

CHECK YOURSELF 4
What is wrong with the field lines in Figure 18.11?

Figure 18.11: For check yourself question 4.

Figure 18.12 shows the electric field lines for two oppositely charged parallel plates. The lines are equally spaced in between the plates indicating that the field there is uniform (i.e. constant). The field stops being uniform at the edges of the plates. The electric field lines are directed from the positive to the negative plate.

Figure 18.12: Field lines for two oppositely charged plates.

The magnitude of the uniform electric field strength in between the plates is given by
E=Vd
where V is the potential difference between the plates and d is the separation of the plates.
CHECK YOURSELF 5
An electron is placed in position A, then B and then C in between two oppositely charged parallel plates (see Figure 18.13).

Figure 18.13: For check yourself question 5.

In which position does the electron experience the greatest force?
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WORKED EXAMPLE 18.4
A proton is placed on the positive plate in a parallel plate arrangement and is then released (see Figure 18.14).

Figure 18.14: For worked example 18.4.

The potential difference between the plates is 120 V and the separation of the plates is 2.0 cm.
aCalculate the electric field strength in between the plates.
bCalculate the force on the proton.
cDetermine the gain in kinetic energy of the proton when it reaches the negative plate.
dThe distance between the plates is doubled but the potential difference remains the same. How do the answers to parts a and b change, if at all?
Answer
aE=Vd=1202.0Ã10â2=6.0Ã103NCâ1.
bF=qE=1.6Ã10â19Ã6.0Ã103=9.6Ã10â16Â N.
cFrom W = qV (review ChapterÂ 11) and W = ÎEK we find ÎEK = 1.6 Ã 10â19 Ã 120 = 1.9 Ã 10â17 J.
dThe electric field strength will halve but the change in kinetic energy will stay the same.
WORKED EXAMPLE 18.5
A proton is placed on the positive plate in a parallel plate arrangement. An electron is placed on the negative plate. Both particles are released at the same time and will reach the opposite plate. The proton is 1840 times more massive than the electron.

Figure 18.15: For worked example 18.5.

aWhat is the ratio gain in kinetic energy of protongain in kinetic energy of electron?
bWhat is the ratio time for proton to reach positive platetime for electron to reach negative plate?
Answer
aThe gain in kinetic energy is qV and so is the same. The ratio is 1:1.
bThe acceleration experienced is a=Fm=qEm and so the time taken is given by (d is the plate separation):
d=12at2=12qEmt2ât=2dmqEââââââ
Hence the required ratio is tptc=mpmcâââââ=1840ââââââ43.
THEORY OF KNOWLEDGE
Action at a distance and fields
These are some of the words of the Scottish theoretical physicist James Clerk Maxwell (1831â1879). Maxwell was one of the scientists who created the concept of the field.
I have preferred to seek an explanation [of electricity and magnetism] by supposing them to be produced by actions which go on in the surrounding medium as well as in the excited bodies, and endeavoring to explain the action between distant bodies without assuming the existence of forces capable of acting directly â¦ The theory I propose may therefore be called a theory of the Electromagnetic field because it has to do with the space in the neighbourhood of the electric and magnetic bodies.
J. C. Maxwell, 1865
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The Millikan oil drop experiment
Between 1910 and 1911, Robert Millikan performed an experiment that measured the charge of an electron. In this experiment, oil drops that were charged by X-rays were allowed to enter a region of uniform electric field between the parallel plates, as shown in Figure 18.16.

Figure 18.16: Schematic diagram of Millikanâs apparatus.

Some oil drops were observed to be at rest in between the plates. This means that the weight of the oil drop is matched by the electric force:
mg=qE
mg=qVd
and so
q=mgdV
So to measure the charge, Millikan needed a method to measure the mass of the oil drop. He did that by turning off the electric field and observed the oil drop that is now falling vertically down.
The oil drop quickly achieves terminal speed because now the weight is matched by the drag force D given by Stokeâs law: D = 6ÏÎ·rv, where Î· is the viscosity of air:
mg=6ÏÎ·rv
So:
r=mg6ÏÎ·v
But the mass of the oil drop is given by m=ÏV=Ï4Ïr33, where Ï is the density of oil.
So:
r=Ï(4Ïr33)g6ÏÎ·v
Giving:
r=9Î·v2Ïgâââââ
Thus, by measuring the terminal speed, Millikan could find the radius of the drop and hence its mass and finally its charge. He measured the charge on more than 150 oil drops but only used the data for 58 oil drops for his published results. (This gave rise to some controversy related to correct scientific practice that you may wish to learn more about.) He found that each charge he measured was an integral multiple of a basic unit; he found that unit to be e = 1.6 Ã 10â19 C.
NATURE OF SCIENCE
Ethics in physics
The case of Robert Millikan played a central role in the discussion of ethics in physics. Millikan was accused of being a misogynist (in a letter to the president of Duke University he advised against hiring women in physics faculty positions) and he expressed anti-Semitic views in letters to his wife while he was in Europe, in contrast to the facts that he personally hired many Jewish physicists at Caltech (including Einstein) and one of his heroes, A. Michelson, was Jewish. He was also accused of mistreating his students: he demanded that a graduate student, who contributed crucially to his experiment, not be an author on the final paper announcing the discovery of the value of the elementary charge.
The big question mark for Millikanâs behaviour is the blatantly false statement in his 1913 paper that âIt is to be remarked, too, that this is not a selected group of drops but represents all of the drops experimented on during 60 consecutive daysâ¦â.
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It is now accepted by all that this is a lie. Millikan only used 58 of the more than 150 oil drops he had measured. So why did Millikan include such a statement in his otherwise masterful paper? Millikan was a perfectionist. Undoubtedly, some runs with oil drops were aborted because something had gone wrong. Others, he may have rejected because he did not trust Stokesâ law for very small drops which were influenced too much by Brownian motion (collisions with air molecules) and others were too big and fell too fast for accurate measurement. Modern historians of science who have gone through Millikanâs notebooks note that even if he had included the data on all the drops, his result for the value of the elementary charge would not change much; only the statistical error would be larger.
TEST YOUR UNDERSTANDING
1A charged conducting sphere with charge +12 ÂµC is allowed to touch an identical conducting sphere of charge â6 ÂµC. The spheres are then separated. What is the charge on each sphere after separation?
2The electroscope is a device consisting of a metal sphere, a metallic rod connected to the sphere and two light metallic leaves attached to the rod. When charge is deposited on the sphere, rod and leaves, the leaves repel as shown. The whole arrangement (see Figure 18.17) is protected within a glass jar.

Figure 18.17: For question 2.

Suppose the charge on the electroscope is positive. A negatively charged rod is put close but not touching the sphere of the electroscope. What will happen to the leaves?
3An electroscope is neutral. The following process takes place:
aa positively charged rod is placed close but not touching the sphere of the electroscope
bthe sphere is earthed
cthe earthing is removed with the charged rod still in place
dthe rod is removed far away.
At each stage draw a diagram to represent the state of the electroscope.
4aCalculate the force between charges q1 of 2.0 ÂµC and q2 of 4.0 ÂµC separated by r = 5.0 cm.
bLet the force calculated in a be F. In terms of F and without further calculations, state the force between these charges when:
ithe separation r of the charges is doubled
iiq1 and r are both doubled
iiiq1, q2 and r are all doubled.
5Three charges are placed on a straight line as shown in Figure 18.18. Calculate the net force on the middle charge.

Figure 18.18: For question 5.

6In the previous question, determine the position of the middle charge so that it is in equilibrium.
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7Calculate the force (magnitude and direction) on the charge q in Figure 18.19, where q = 3.0 ÂµC.

Figure 18.19: For question 7.

8A charge of magnitude +5.0 ÂµC experiences an electric force of magnitude 3.0 Ã 10â5 N when placed at a point in space. Determine the electric field at that point.
9Figure 18.20 shows two spherical charges that are fixed in place. The spheres have the same radius.

Figure 18.20: For question 9.

Explain:
ain which region, I, II or III, could the electric field be zero?
bin which region, I, II or III, does the electric field attain its maximum magnitude?
10Four charges, each of magnitude q, are fixed at the vertices of a square of side a, as shown in Figure 18.21.

Figure 18.21: For question 10.

aWhat is the direction of the electric field at the centre of the square?
bWhat is the magnitude of the field in a?
11Two plastic spheres each of mass 100.0 mg are suspended from very fine insulating strings of length 85.0 cm, as shown in Figure 18.22.
When equal positive charges are placed on the spheres, the spheres repel and are in equilibrium when 10.0 cm apart.

Figure 18.22: For question 11.

aDetermine the charge on each sphere.
bEstimate how many electron charges this corresponds to.
cThe charge on the left sphere is increased. Draw a diagram to show how the two charges hang.
12Consider two people, each of mass 60 kg, a distance of 10 m apart.
aAssuming that all the mass in each person is made out of water, estimate how many electrons there are in each person.
bHence, estimate the electrostatic force of repulsion between the two people due to the electrons.
cList any other simplifying assumptions you have made to make your estimate possible.
dNo such force is observed in practice. Suggest why this is so.
13The electric field is a vector, and so two electric fields at the same point in space must be added according to the laws of vector addition. Consider two equal positive charges q, each 2.00 ÂµC, separated by a = 10.0 cm and a point P a distance of d = 30.0 cm, as shown in Figure 18.23.
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Figure 18.23 shows the directions of the electric fields produced at P by each charge. Determine the magnitude and direction of the net electric field at P.

Figure 18.23: For question 13.

14Repeat the previous question where the top charge is +2.00 ÂµC and the bottom charge is â2.00 ÂµC.
15The formula for the radius of the oil drop derived in the text ignores the buoyant force on the oil drop. Show that if this force is taken into account the radius is given by r=9Î·v2(ÏâÏai)gâââââââââââ.
16In a hypothetical Millikan type experiment the following charges were measured on hypothetical oil drops: 96 Ã 10â20 C, 144 Ã 10â20 C, 192 Ã 10â20 C, 240 Ã 10â20 C, 288 Ã 10â20 C. What is the unit of electric charge in this âexperimentâ? Explain your reasoning.
