Electric Potential: Definition, Formula, Units, and Solved Problems

 17.4 ELECTRIC POTENTIAL

17.4.1 Work Done on Charge

Figure 17.32

1. By external force:

(a) Consider a positive point charge q at point A in a uniform electric field which points to the right, as shown in Figure 17.32(a). An electric force $F_e$ acts on the charge and it too points to the right.

(b) Because of this, an external force F has to be applied on the charge in order to move the charge from point A to another point B to the left. To move the charge we must have

$F \geqslant -F_e$

2. By electric force

(a) The work $W_{AB}$ done by the electric force is given by

$W_{AB}=F_e \Delta x \ cos \ 180^0$

$W_{AB}=-F_e \Delta x$ (negative)

(b) If we let 

$F'=F_e$

then we have 

$W_{AB}=-W_{AB}'$

Take note of the following points:

(a) $W_{AB}$​ is the work done on the electric field by the external force.

(b) The work done by the electric force in moving a positive charge in the direction opposite to the electric field is negative.

(c) The work $W_{AB}'$ done by the external force is converted to electric potential energy. As a result, the charge system gains energy. This is analogous to a person lifting an object upwards, hence doing work against the gravitational pull. The object, now at greater height than before, has gained greater gravitational potential energy.

EXAMPLE 17.11

A point charge $q = +2.0\,\mu C$ is placed at point A in a uniform electric field of strength $5.0\,kN\,C^{-1}$. An external force moves the charge at constant speed in a direction opposite to the direction of the field to point B, $5.0\,cm$ away. Determine the work done on the charge at the end of the transfer

(a) by the external force
(b) by the electric force

Answer

(a) The electric force acting on $q$ is given by

$F_e=qE=(2 \times 10^{-6})(5\times 10^3)=1.0 \times 10^{-2}$ N

Since the charge moves at constant speed, the magnitude of the external force is

$F'=F_e=1.0 \times 10^{-2}$ N

Work done by the external force is

$W_{AB}'=F' \Delta x=(1.0 \times 10^{-2})(0.05)=+0.50 mJ$

(b) Work done by the electric force is

$W_{AB}=-W_{AB}'=-0.05 mJ$

At the end of the charge transfer, the charge system gains electric PE of 0.50 mJ.

17.4.2 Electric Potential Energy

Electric force is a conservative force. Because of this, any work that is done by an electric force can be associated with potential energy, in this case electric potential energy. Refer to Figure 17.32(a). The positive charge q stores electric potential energy $U_A$ when it is at point A. While it is at point B it stores electric potential energy $U_B$. This is analogous to a mass storing gravitational PE at a point at height $h$ above the ground in the gravitational field.

17.4.3 Path Taken from One Point to Another in Electric Field

Since electric force is a conservative force, the work done in transferring a charge from point A to point X via path $C_1$ is the same as that via path $C_2$, as shown in Figure 17.33. The work is independent of the path taken. The work is dependent upon the position of the final destination, like point B.

Figure 17.33

17.4.4 Work Done by External Force in Transferring Positive Charge at Infinity into an Electric Field

1. A positive point charge $q$ is initially at infinity. We intend to bring this charge into an electric field which is produced by a positively charged system. Suppose an external force F' acts on $q$ and pushes $q$ to a point A in the electric field. The direction of motion of $q$ is opposite to the direction of the electric field. Because of this, a repulsive electric force acts on $q$ when it is in the electric field. This is analogous to a mass being lifted vertically upwards by an external force in the gravitational field.

2. Work has to be done by the external force F' in order to overcome the repulsive electric force. When the charge has reached point A, let the work done on $q$ by F' be W∞→A′​. This work done by the external force is converted to electric potential energy. Hence, charge $q$ at A has gained electric potential energy, and the amount gained, $\Delta U_{\infty \rightarrow A}$, is given by

$$W'_{\infty \rightarrow A} = +\Delta U_{\infty \rightarrow A}$$

where$\Delta U_{\infty \rightarrow A}$ is the difference between the electric potential energy $U_A$ stored by $q$ at A and the electric potential energy $U_\infty$ stored by $q$ at infinity. In other words,

$$\Delta U_{\infty \rightarrow A} = U_A - U_\infty$$

Hence

$$W'_{\infty \rightarrow A} = +(U_A - U_\infty)$$

3. By convention, the electric potential energy $U_\infty$ stored by a charge at infinity is assigned a value of zero, i.e.,

$$U_\infty = 0$$

We can rewrite the expression for work done $\Delta W'_{\infty \rightarrow A}$ by F' as

$$W'_{\infty \rightarrow A} = +U_A$$

This expression merely states that the energy utilized in doing work$(W'_{\infty \rightarrow A})$ is finally converted to electric potential energy stored by the charge system.

17.4.5 Work Done by External Force in Transferring Positive Charge from One Point to Another in Electric Field

1. Suppose the external force $F'$ pushes charge $q$, which is now at point A, further into the electric field to point B, moving in a direction opposite to the electric force (Figure 17.29(a)). The charge $q$ stores electric potential energy $+U_B$ when it is at B. This electric PE is equal to the work $W'_{\infty \rightarrow B}$ done on $q$ by the external force in bringing $q$ from infinity to point B. In other words,

$$W'_{\infty \rightarrow B} = +U_B$$

2. The external force has to pass through A and travel further into the electric field in order to reach B. Hence, the work done by the external force in bringing $q$ at infinity to B is more than that in bringing $q$ at infinity to A. Because of this, we have

$$U_B > U_A$$

3. Hence, the amount of work $W'_{AB}$ done on $q$ by the external force in bringing $q$ from A to B is given by

$$W'_{AB} = W'_{\infty \rightarrow B} - W'_{\infty \rightarrow A}$$$$= (+U_B) - (+U_A)$$
$$= +\Delta U_{AB}$$

where

$$\Delta U_{AB} = U_B - U_A$$

In other words, the charge system gains electric PE by an amount $\Delta U_{AB}$ due to work done by the external force that brings the positive charge from one point in the electric field to another point in a direction that is opposite to the direction of the electric force.

17.4.6 Work Done by Electric Force in Transferring Positive Charge from One Point to Another in Electric Field

Refer to Figure 17.31(a). An electric force $F_e$ acts on $q$ when the charge is at A. When $q$ is brought to B, work $W_{AB}$ is done by the electric force. We have

$$W_{AB} = -F_e \Delta x$$

(from A to B)

$$= -W'_{AB}$$$$W_{AB} = -\Delta U_{AB}$$

Summary

We have the following when a positive charge moves from one point to another in an electric field:

Charge moving in the same direction as the electric force	Charge moving in the direction opposite to the electric force
(1) Work is done by the field.	Work is done on the field.
(2) Work done by (Fe) is positive.	Work done by (Fe) is negative.
(3) Charge loses electric PE. (ΔU) is negative (U decreases).	Charge gains electric PE. (ΔU) is positive (U increases).

17.4.7 Electric Potential at a Point

1. Meaning

Refer to Figure 17.32(a). The point charge of +q coulomb at point A has electric potential energy $U_A$. Then the electric PE of a charge of +1 C at A is given by $U_A/q$. This quantity of electric PE stored by 1 C of charge is known as the electric potential at a point in the electric field. It is represented by the letter $V$. Hence, the electric potential at point A is

$$V_A = \frac{U_A}{q}$$

2. Definition

Suppose a positive charge $q$ is transferred from infinity to point A in an electric field produced by a positively charged object. The work $W_{\infty \rightarrow A}$ done on charge $q$ by the electric force is given by

$$W_{\infty \rightarrow A} = -\Delta U$$
=−(UA−U∞)=−UA(U∞=0)

Hence

$$V_A = \frac{W_{\infty \rightarrow A}}{q}$$

If $q = +1\,C$, then $V_A=\mathrm{\mid }{W}_{\mathrm{\infty }\to A}\mathrm{\mid .\; Based\; on\; this\; equation,\; we\; have\; the\; following\; definition:}$

The electric potential at a point in an electric field is the energy required to transfer 1 C of positive charge from infinity to that point.

3. Quantity

Electric potential is a scalar quantity.

4. Unit

Electric potential has a unit of J C⁻¹ or volt (V).

5. Electric potential drop or rise in an electric field

The electric potential value

(a) drops as we move a positive charge from one point to another point in the same direction as the electric field lines

(b) rises as we move a positive charge from one point to another point in the direction opposite to the electric field lines

6. Electric potential energy at a point

If the electric potential at a point in an electric field is V, then a charge $q$ at that point has electric potential energy U of amount

$$$$U = qV 

EXAMPLE 17.12

A point charge $q=+5.0\mu \mathrm{C }$at infinity is brought to a point X in an electric field which is produced by a positively charged object. Work of magnitude 30 mJ is done on the field in the transfer of this charge.

(a) Determine the electric PE stored by the charge at X.
(b) Determine the electric potential at X.
(c) The positive charge is removed. A charge q′ = −1.5μC is placed there. Determine the electric PE stored at X.

Answer

(a) Since the electric field is produced by a positively charged object, charge q will experience a repulsive electric force when it is brought into the field. Hence, an external force has to act on q in order to bring it from infinity to point X. The work 

$W_{\infty-X}=-W'_{\infty-X}=-30 \ mJ$

But 

$W_{\infty-X}=-\Delta U_{\infty-X}$

$W_{\infty-X}=-(U_X-U_{\infty})=-U_X$

Electric PE, $U_X=-W'_{\infty-X}=-(-30 \ mJ)=+30 \ mJ$

(b) Electric potential, $V=\frac{U_X}{q}$

$V=\frac{+30 \ mJ}{+ 5 \mu C}=+6.0 \ kV$

(c) Electric PE, $U'_X=q'V_X$

$U'_X=(-1.5 \mu C)(+6.0 \ kV)=-9.0 \ mJ$

17.4.8 Electric Potential Potential Difference

1. Meaning

Refer fo Figure 17.32. Electric potential $V_A$ and $V_B$ exist at points A and B respectively in the electric field. When charge +q travels from A to B (or B to A), it will experience a change in electric potential, from $V_A$ to $V_B$ (or $V_B$ to $V_A$). In other words, there exist a difference in electric potential across A and B. This difference in the electric potential across two points in an electric field is known as the electric potential difference (p.d).

2. Work and p.d

The electric potentials at points A and B are given by

$V_A=-\frac{W_{\infty-A}}{q}$

$V_B=-\frac{W_{\infty-B}}{q}$

or $W_{\infty-A}=qV_A$

$W_{\infty-B}=qV_B$

Work $W_{AB}$ is done on q by the electric force in transferring the charge from A to B. We have

$W_{AB}=W_{\infty-B}-W_{\infty-A}$

$W_{AB}=(-qV_B)-(-qV_A)$

$=(-qV_B)-(-qV_A)$

$=-q(V_B-V_A)$

$=-q \Delta V_{AB}$

where $\Delta V_{AB}=V_B-V_A$ is the electric p.d. across A and B. In general, we have

$W=-q \Delta V$

3. Definition of electric p.d.

We have

$W_{AB}=-q \Delta V_{AB}$

Let q = +1 C, then $\Delta V_{AB}=-W_{AB}$ = work done on field if there is a potential rise. Hence the electric potential difference between points A and B in an electric field is defined as:

the energy required to transfer 1 C of positive charge from point A to point B.

EXAMPLE 17.12

The electric potentials at points X and Y in an electric field are given below:

A point charge q = +8.0 μC is transferred from X to Y.
(a) Determine the work done in this charge tranfer
(b) Is work done on or by the electric field?
(c) Does the charge gain or lose electric PE at the end od the transfer?

Answer

For (p) (a) Work done $W_{XY}=-q(V_Y-V_X)$

$=-(+8 \times 10^{-6})[(-200)-(+100)]$

$=-(+8 \times 10^{-6})(-300)=+2.4 \times 10^{-3} \ J$

Note: $\Delta V_{XY}=-300 \ V$ is negative. The charge passes through a potential drop.

(b) Since $W_{XY}$ is positive, work is done by the electric field.

(c) Change in electric PE $\Delta U_{XY}=-W_{XY}$ 

$=-(+2.4 \ mJ)=-2.4 \ mJ$

Since $\Delta U_XY$ is negative, there is a decrease in electric PE. The charge has lost electric PE of 2.4 mJ.

For (q) (a) Work done $W_{XY}=-q(V_Y-V_X)$

$=-(+8 \times 10^{-6})[(+5)-(+3)] \times 10^3$

$=-16 \ mJ$

Note: $\Delta V_{XY}=+2 \ kV$ is positive. The charge passes through a potential rise.

(b) Since $W_{XY}$ is negative, work is done on the field.

(c) Change in electric PE $\Delta U_{XY}=-W_{XY}$ 

$=-(-16 \ mJ)=+16 \ mJ$

Since $\Delta U_XY$ is positive, there is a increase in electric PE. The charge has gained electric PE of 16 mJ

 EXAMPLE 17.14

A point charge $\mathrm{<br>}$ travels from point X to point Y in an electric field crossing a potential rise of 200 V.

(a) Determine the work done in transferring the charge.
(b) Is work done on or by the electric field?
(c) An electric potential of −20 V exists at point Y. Determine the electric potential at point X.

Answer

(a) $W_{XY}=-q \Delta V_{XY}$

$=-(+2 \times 10^{-6})(+200)=-0.40 \ mJ$

(b) Since the work is negative, work is done on the electric field.

Note: A potential rise already implies that work is done on the field.

(c) $W_{XY}=-q(V_Y-V_X)$

$-0.40 \times 10^{-3}=-(+2 \times 10^{-6})[(-20-V_X)]$

$+200 = -20-V_X$

$V_X=220 \ V$

OR

p.d. rise

$\Delta V_{XY}=V_Y-V_X=+200 \ V$

$V_X=V_Y-200$

$V_X=(-20)-200 =-220 \ V$

EXAMPLE 17.15

A negative point charge of 0.75 mC travelling from point X to point Y in an electric field experiences a p.d. drop of 200 V.

(a) Determine the potential at Y if a potential of +20 V exists at point X.
(b) Determine the work done in this charge transfer. Is this work done on or by the field?
(c) Determine the electric PE at Y. Has the charge gained or lost electric PE when it is at Y? How much has it gained or lost?

Answer

(a) Potential drop

$\Delta V_{XY}=V_Y-V_X=-200$

$V_X=V_Y+200$

$V_X=(+20)+200=+220 \ V$

(b) Since the charge is negative and experiences a potential drop, it means that work is done on the field. 

$W_{XY}$ has to be negative.

$W_{XY}=-q\Delta V_{XY}$

$=-(-0.75 \times 10^{-3})(-200)=-0.15 \ J$

(c) Electric PE at Y

$U_y=qV_y$

$=(-0.75 \times 10^{-3})(+20)=-0.015 \ J$

Since work is done on the electric field, the charge gains electric PE when it reaches point Y.       

electric PE gained $=\; \$\backslash Delta\; U\_\{XY\}\$$

$W_{XY}=-(-0.15)=+0.15 \ J$

Note: At X,

$U_X=qV_X$

$U_X=(-0.75 \times 10^{-3})(+220)=0.165 \ J$

$U_X < U_Y$

​Hence, there is a gain in electric PE as the negative charge moves from X to Y.

EXAMPLE 17.16

Work of magnitude 0.20 mJ is done on an electric field when a charged particle travelling from a point X to another point Y in the field, experiences an electric potential drop of 50 V. Determine the charge carried by the particle.

Answer

Work done on the field is given by

$W_{XY}=-q \Delta V_{XY}$

$-0.2 \times 10^{-3}=-q(-50)$

$q=-4.0 \mu C$

17.4.9 Electric Potential at a Point in an Electric Field Produced by an Isolated Point Charge

1. Formula for V at a point

A positive point charge q is at infinity. It is then moved towards a fixed isolated positive point charge $$. At some particular instant, q has reached a point which is at a distance of x from $$. The electric force F acting on q at that point is

$F=k\frac{Qq}{x^2}$

Charge q is then forced to move slightly towards Q through a distance of $\mathrm{<br>}$ by an agent. The work done by the agent in overcoming force $$ is

$\Delta W'=(F)(\Delta x)(cos \ 180^0)$

$=-k\frac{Qq}{x^2}\Delta x$

In order to bring q from infinity to a point A at a distance of r from Q, the total work $W'_{\infty-A}$ done is

$W'_{\infty-A}=\int_{}^{}dW$

$=-kQq\int_{\infty}^{r}x^{-2}dx$

$=-kQq \left[ -x^{-1} \right]_{\infty}^r$

$=k\frac{Qq}{r}$

But, $W'_{\infty-A}$= gain in electric potential energy by system = $qV_A$

$V=\frac{W'_{\infty-A}}{q}$

$V=k\frac{Q}{r}$

or, $V=\frac{Q}{4\pi \epsilon_0r}$

This equation is for a point in the electric field produced by a point charge in free space.

2. Sign of V

The sign of V determined by the sign of Q. If Q is postive, then V is positive. Conversely, if Q is negative, then V is negative.

3. Graph of V against r

We have 

$V=\frac{Q}{4\pi \epsilon_0r}$

$V\propto \frac{1}{r}$   (Q constant)

Based on this relation, we have graphs of V against r as shown in Figure 17.33(a) and (b)

EXAMPLE 17.17

The electric potential at a point at certain distance from a point charge is +600 V. The electric field strength at that point has a magnitude of 3.0 kV/m. Determine

(a) the distance of the point from the point charge
(b) the charge of the point charge.

Answer

(a) We have, $V=k\frac{Q}{r}$ and $E=k\frac{Q}{r^2}$

Thus, $r=\frac{V}{E}$

$r=\frac{600 \ V}{3000 \ V/m}=0.20 \ m$

(b) $V=k\frac{Q}{r}$

$+600 = (9 \times 10^9)\left(\frac{Q}{0.2}\right)$

$Q=+1.3 \times 10^{-8} \ C=13.0 \mu C$

EXAMPLE 17.18

Points A and B are at distances 2.0 cm and 3.0 cm respectively from a point charge
$Q=-100\mu C$. Determine

(a) the electric potentials at A and B
(b) the energy required in moving a point charge $q=+2.0\mu C$, which is placed at A, to B.

Answer

(a) $V_A=k\frac{Q}{r_A}$

$V_A=(9 \times 10^9)\frac{-100 \times 10^{-6}}{2 \times 10^{-2}}$

$V_A=-4.5 \times 10^7 \ V$

$V_B=k\frac{Q}{r_B}$

$V_A=(9 \times 10^9)\frac{-100 \times 10^{-6}}{3 \times 10^{-2}}$

$V_A=-3 \times 10^7 \ V$

(b) The energy required is given by

$W_{AB}=-q\Delta V_{AB}$

$=-q(V_B-V_A)$

$=-(+2 \times 10^{-6})[(-3.0)-(-4.5)]\times 10^7$

$=-30 \ J$

Note

(i) When the positive charge $q$ moves from A to B, away from $-Q$, work is done on the field and the amount of work done is 30 J.

(ii) $W_{AB}=\Delta U_{AB}=-(-30)=+30 \ J$. Hence, experiences a potential rise as it moves from A to B.

EXAMPLE 17.19

A point X is at a distance of 50 cm from another point charge $Q=+10\mu C$. A point Y lies on the line joining X and Q and is 20 cm from Q. Determine

(a) the work done by an external force in bringing a point charge q = +1.5 μC at infinity to point X 
(b) the work done by the electric force in moving q from X to Y

Answer

(a) The electric potential at X due to Q is given by

$V_X=k\frac{Q}{r_X}$

$=(9\times 10^9)\left(\frac{+10\times 10^{-6}}{0.50}\right)=+180 \ kV$

The work done by the external force in bringing q at infinity to point X is given by

$W'_{\infty-X}=+\Delta U_{\infty-X}$

$=+(U_X-U_{\infty})$

$=+(qV_X-0)$

$=+(+1.5 \times 10^{-6})(+180 \times 10^3)$

$=+0.27 \ J$

(b) The electric potential at Y due to Q is given by

$V_Y=k\frac{Q}{r_Y}$

$=(9\times 10^9)\left(\frac{+10\times 10^{-6}}{0.20}\right)=+450 \ kV$

The work done by the external force in bringing q at infinity to point X is given by

$W'_{\infty-X}=-q \Delta V_{XY}$

$=+q(V_Y-V_X)$

$=-(+1.5 \times 10^{-6})[(+450)-(+180)]\times 10^3$

$=-0.41 \ J$

17.4.10 Resultant Electric Potential at a Point in Electric Field Produced by Several Point Charges

Consider a point X inside the resultant electric field formed by the superposition of the fields produced by two point charges $Q_1$ and $Q_2$ ​, as shown in Figure 17.34. Let the electric potential at X due to $Q_1$  alone be $V_1$ ​ and due to $Q_2$  alone be $V_2$ ​. Since electric potential is a scalar quantity, the resultant electric potential $V_X$ ​ at X is given by

$${\mathrm{\$V\_X=V\_1+V\_2\$}}_{}$$

In general, if we have n point charges, then the resultant potential will be

$${\mathrm{\$V\_X=V\_1+V\_2+...+V\_n\$}}_{\mathrm{<br>}}$$

EXAMPLE 17.20

A point charge $Q_1$ = +5.0 μC and another point charge $Q_2$ ​= −4.0 μC are separated by a distance of 30 cm. A point X lies on the straight line joining $Q_1$​ and $Q_2$​ and is 10 cm from $Q_1$​. Determine the work done on or by the electric field in moving a point charge $q=+0.50\mu C$ from infinity to point X. At the end of the charge transfer, does the charge system gain or lose electric PE?

Answer

At point X, the resultant electric potential is given by

$V_X$ = potential due to $Q_1$ + potential due to $Q_2$

$=k\left(\frac{Q_1}{r_1}+\frac{Q_2}{r_2}\right)$

$=(9 \times 10^9)\left(\frac{+5}{0.10}+\frac{-4}{0.20}\right)$

$=+2.7 \times 10^5 \ V$

Work done on or by the electric field in bringing charge q from infinity to point X is given by

$W_{\infty-X}=-q(V_X-V_{\infty})$

$=-(+0.5 \times 10^{-6})[(+2.7 \times 10^5)-0]=-0.14 \ J$

Since the work done is negative, work is done on the field in bringing charge q from infinity to point X. The electric PE of the charge system increases. This energy gained by the field is given by an agent, which transfers charge q from infinity to X. Hence the work done by the agent is

$W'_{\infty-X}=-W_{\infty-X}=+0.14 \ J$

EXAMPLE 17.21

Point charges are placed at each corner of a square of length 10 cm, as shown in Figure 17.35. Determine the work done in moving a point charge $q=-2.0\mu C$ from the centre O of the square to the midpoint E on AD. At the end of the charge transfer, does the charge system gain or lose electric PE?

Answer

At O, the resultant electric potential is given by

$$V_O = V_1 + V_2 + V_3 + V_4$$ $$= k \left(\frac{Q_1}{r} + \frac{Q_2}{r} + \frac{Q_3}{r} + \frac{Q_4}{r}\right)$$
$$r = \frac{1}{2}\sqrt{0.10^2 + 0.10^2} = 0.0707 \, m$$
$$V_O = \frac{(9 \times 10^9)}{0.0707} [(+0.010)+(+0.020)+(-0.020)+(+0.010)] \times 10^{-6}$$
$$= +2546 \, V$$

At E, the resultant electric potential is given by

$$V_E = k \left(\frac{Q_1}{r_1} + \frac{Q_2}{r_2} + \frac{Q_3}{r_3} + \frac{Q_4}{r_4}\right)$$

We have

$$r_1 = AE = 0.050 \, m$$
$$r_2 = BE$$
$$r_3 = CE$$
$$r_4 = DE = 0.050 \, m$$
$$V_E = (9 \times 10^9) \left( \frac{+0.01}{0.050} + \frac{+0.02}{BE} + \frac{-0.02}{CE} + \frac{+0.01}{0.050} \right) \times 10^{-6}$$
$$= +3600 \, V$$

Work done on or by the electric field in moving charge $q = -2.0 \, \mu C$ from O to E is given by

$$W_{OE} = -q(V_E - V_O)$$
$$= -(-2 \times 10^{-6})[(+3600) - (+2546)]$$
$$= +2.1 \times 10^{-3} \, J$$

Since the value is positive, work is done by the electric field. The charge system loses electric PE at the end of the charge transfer.

In this case an agent is not needed to move charge q from O to E. The energy needed in the transfer is obtained from the electric PE of the charge system. This is analogous to a mass falling freely under gravity.

 

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