17.4 ELECTRIC POTENTIAL
17.4.1 Work Done on Charge
Figure 17.32 A positive charge in a uniform electric field
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 Fe 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
$F ≥ −F_e$
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) WAB 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 μC 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 qqq is given by
Fe = qE = (2 × 10⁻⁶)(5 × 10³) = 1.0 × 10⁻² N
Since the charge moves at constant speed, the magnitude of the external force is
F' = Fe = 1.0 × 10⁻² N
Work done by the external force is
W'AB = F' Δx = (1.0 × 10⁻²)(0.05) = +0.50 mJ
(b) Work done by the electric force is
WAB = −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 UA when it is at point A. While it is at point B it stores electric potential energy UB.
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 C₁ is the same as that via path C₂, 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 Since electric force is a conservative force, the work done in transferring a charge from point A to point X via path C₁ is the same as that via path C₂
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 $W'_{\infty \to A}$. 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, $\Delta U_{\infty \to A}$, is given by
$W'_{\infty \to A}=\Delta U_{\infty \to A}$
where $\Delta U_{\infty \to A}$ 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,
$\Delta U_{\infty \to A}=U_A-U_{\infty}$
Hence
$W'_{\infty \to A}=U_A-U_{\infty}$
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 $W'_{\infty \to A}$ by F' as
$W'_{\infty \to A}=U_A$
This expression merely states that the energy utilized in doing work $(W'_{\infty \to A})$ 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 $W'_{\infty \to B}$ done on q by the external force in bringing q from infinity to point B. In other words,
$W'_{\infty \to A}=\Delta U_{\infty \to A}$
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
$W'_{AB}=W'_{\infty \to B}-W'_{\infty \to A}$
$W'_{AB}=(+U_B)-(+U_A)$
$W'_{AB}=\Delta U_{AB}$
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 Fe 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)
$\boxed{W_{AB}=\Delta U_{AB}}$
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 $\frac{U_A}{q}$. 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 qqq is transferred from infinity to point A in an electric field produced by a positively charged object. The work $W_{\infty -A}$ done on charge qqq by the electric force is given by
$W_{\infty -A}= - \Delta U$
$W_{\infty -A}= -(U_A-U_{\infty})=-U_A$ ($U_{\infty}=0$)
Hence
$V_A=\frac{W_{\infty \to A}}{q}$
If q = +1 C, then $V_A=∣W_{\infty \to A}∣$. 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 qq at that point has electric potential energy U of amount
$\boxed{U = qV}$
Example 17.12
A point charge q = +5.0 μC 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′ = −1.5 μC 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 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.
$W_{AB}=-q \Delta V_{AB}$
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 17.13 – Work Done in Electric Potential Difference
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.
| Vx | Vy |
|---|---|
| (p) +100 V | −200 V |
| (q) +3.0 kV | +5.0 kV |
(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 – Work Done and Electric Potential
A point charge q = +2.0 μC 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 – Electric Potential Energy
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 =$\Delta U_{XY}$
$W_{XY}=-(-0.15)=+0.15 \ J$
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 Q. At some particular instant, q has reached a point which is at a distance of x from Q. 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 Δx by an agent. The work done by the agent in overcoming force F 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
$W'_{\infty-A}=\int_{}^{}dW$
$=-kQq\int_{\infty}^{r}x^{-2}dx$
$=-kQq \left[ -x^{-1} \right]^r_{\infty}$
$=k\frac{Qq}{r}$
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 μC. Determine
(a) the electric potentials at A and B
(b) the energy required in moving a point charge q = +2.0 μC, 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 μC. 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
$V_X = V_1 + V_2$
In general, if we have n point charges, then the resultant potential will be
$V_X = V_1 + V_2 + ... + V_n$
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 μC 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 μC 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
$V_O = V_1 + V_2 + V_3 + V_4$
$V_O = k\left(\frac{Q_1}{r}+\frac{Q_2}{r}+\frac{Q_3}{r}+\frac{Q_4}{r}\right)$
$r=\frac{1}{2}\sqrt{0.10^2+0.10^2}=0.0707 \ m$
$V_O = \left(\frac{9 \times 10^9}{0.0707}\right)[(+0.010)+(+0.020)+(-0.020)+(+0.010)] \times 10^{-6}$
$V_0= +2546 \ V$
At E, the resultant electric potential is given by
$V_E = k\left(\frac{Q_1}{r_1}+\frac{Q_2}{r_2}+\frac{Q_3}{r_3}+\frac{Q_4}{r_2}\right)$
We have
$r_1=AE=0.050 \ m$, $r_2=BE$, $r_3=CE$, $r_4=DE=0.050 \ m$
$V_E=(9 \times 10^9)\left(\frac{+0.01}{0.050}+\frac{+0.02}{BE}+\frac{-0.02}{CE}+\frac{+0.01}{0.050}\right)\times10^{-6}$
$V=+3600 \ V$
Work done on or by the electric field in moving charge q = −2.0 μC from O to E is given by
$W_{OE}=-q(V_E-V_O)$
$W=-(-2 \times 10^{-6})[(+3600)-(+2546)]$
$W=+2.1 \times 10^{-3} \ J$
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.
EXAMPLE 17.22
|
| Figure 17.36 Point charges $Q_1 = +5.0 \mu C$ and $Q_2 = −10 \mu C$ are placed at points A and B respectively |
Point charges $Q_1 = +5.0 \mu C$ and $Q_2 = −10 \mu C$ are placed at points A and B respectively, as shown in Figure 17.36. C = 5.0 m and CB = 4.0 m. Determine:
(a) the resultant electric potential at point C
(b) the work required in transferring a point charge $q = −0.050 \mu C$ from infinity to C at constant speed. Does the charge system gain or lose energy at the end of the transfer?
Answer
(a) $V_C = k\left(\frac{Q_1}{AC}+\frac{Q_2}{CB}\right)$
$V_C=(9 \times 10^9)\left(\frac{+5}{5}+\frac{-10}{4}\right)\times10^{-6}$
$V_C=-13.5 \times 10^3 \ V$
(b) $W_{\infty \to C}=-q(V_C-V_\infty)$
$W_{\infty \to C}=-(-0.05 \times 10^{-6})[(-13.5\times10^3)-0]$
$W_{\infty \to C}=-0.68 \ mJ$
Since the value is negative, work is done on the electric field. The charge system gains electric PE.
The energy gained is obtained from the agent that transfers the charge. The agent has to do work against the net repulsive electric force acting on qqq.
The work done by the agent is
$W'_{\infty \to C}=-W_{\infty \to C}$
$W'_{\infty \to C}=+0.68 \ mJ$
EXAMPLE 17.23
|
| Figure 17.37 A rectangle ABCD has length AB = 0.40 m and breadth BC = 0.30 m |
A rectangle ABCD has length AB = 0.40 m and breadth BC = 0.30 m, as shown in Figure 17.37. Point charges $Q_1= +5.0 \ \mu C$ and $Q_2 = -10 \ \mu C$ are placed at corners A and C respectively. Determine:
(a) the electric p.d. $V_D - V_B$
(b) the work required in transferring a point charge $q = +0.10 \ \mu C$ from B to D. Does the charge system gain or lose electric PE at the end of the transfer?
Answer
(a) $V_B = k \left(\frac{Q_1}{AB} + \frac{Q_2}{BC}\right)$
$V_B = (9 \times 10^9) \left( \frac{+5}{0.4} + \frac{-10}{0.3} \right)\times10^{-6}$
$V_B= -188 \times 10^3 \ V$
$V_D = k \left(\frac{Q_1}{AD} + \frac{Q_2}{CD}\right)$
$V_D= (9 \times 10^9) \left( \frac{+5}{0.3} + \frac{-10}{0.4} \right)\times10^{-6}$
$V_D - V_B= (-75 \times 10^3) - (-188 \times 10^3)$
$V_D - V_B= +113 \ kV$
(b) $W_{BD} = -q(V_D - V_B)$
$W_{BD}=-(+0.10 \times 10^{-6})(+113 \times 10^3)$
$W_{BD}= -11 \ mJ$
Since the value is negative, work is done on the electric field. The charge system will gain electric PE at the end of the transfer.
The energy gained by the charge system comes from the agent that does work against the repulsive electric force acting on qq. The work done by the agent is +11 mJ.
17.4.11 Conclusion: Electric Potential
Electric potential is a fundamental concept in electrostatics that describes the electric potential energy per unit charge at a point in an electric field. It helps us understand how electric charges store and transfer energy within an electric field.
Electric potential is a scalar quantity and is measured in volts (V), where 1 V = 1 J/C. The electric potential at a point depends on the source charge and the distance from that charge, and for a point charge it is given by:
$V = k\frac{Q}{r}$
Electric potential is closely related to electric potential energy, where the energy of a charge in an electric field is expressed as:
$U = qV$
When a charge moves between two points with different potentials, work is done either by the electric field or by an external force. The work involved is related to the potential difference between the points.
In systems with multiple charges, the principle of superposition applies, meaning the total electric potential at a point is the sum of the potentials produced by all charges.
Overall, electric potential provides an important way to analyze energy changes, work, and interactions of charges in electric fields, making it a key concept in understanding many electrical and physical phenomena.
17.4.12 Applications of Electric Potential
Electric potential has many important applications in physics, engineering, and everyday technology.
1. Electric Circuits
Electric potential difference (voltage) drives the flow of electric current in circuits. Devices such as batteries provide a potential difference that allows electrical energy to power appliances and electronic devices.
2. Capacitors
Capacitors store electrical energy using electric potential. They are widely used in electronic circuits, power supplies, cameras, and energy storage systems.
3. Electron Accelerators
Electric potential is used to accelerate charged particles such as electrons and protons. This principle is applied in particle accelerators, cathode ray tubes, and some medical equipment.
4. Electrostatic Precipitators
Industries use electric potential to remove dust and smoke particles from exhaust gases. Charged particles are attracted to plates with opposite electric potential, helping reduce air pollution.
5. Photocopiers and Laser Printers
These devices use electric potential to control charged toner particles so that they attach to paper and form images or text.
6. Medical Applications
Electric potential is used in medical technologies such as electrocardiograms (ECG) and electroencephalograms (EEG), which measure electric potentials generated by the heart and brain.
7. Energy Storage Systems
Electric potential is important in batteries and supercapacitors, where electrical energy is stored and later supplied to devices.
8. Particle and Ion Movement
Electric potential controls the motion of charged particles in devices like mass spectrometers and ion traps used in scientific research.
9. Electric Field Mapping
Engineers use electric potential to study electric fields and design electrical equipment such as transformers and insulators.
10. Plasma and Space Physics
Electric potential helps explain the behavior of charged particles in plasmas, lightning, and phenomena in space environments.
17.4.13 Frequently Asked Questions (FAQ) – Electric Potential
1. What is electric potential?
Electric potential is the electric potential energy per unit charge at a point in an electric field. It tells how much energy a charge would have at that position.
$V = \frac{U}{q}$
where
V = electric potential (volt), U = electric potential energy (joule), q = charge (coulomb)
2. What is the unit of electric potential?
The SI unit of electric potential is the volt (V).
$1 \ V = \frac{J}{C}$
3. Is electric potential a scalar or vector quantity?
Electric potential is a scalar quantity, meaning it has magnitude only and no direction.
4. What is the electric potential due to a point charge?
The electric potential produced by a point charge is given by:
$V = k\frac{Q}{r}$
where
k = 9 × 109 N·m²/C², Q = source charge, r = distance from the charge
5. What is the difference between electric potential and electric potential energy?
| Electric Potential | Electric Potential Energy |
|---|---|
| Energy per unit charge | Energy stored by a charge |
| Unit: Volt (V) | Unit: Joule (J) |
| (V = U/q) | (U = qV) |
6. What happens to electric potential along electric field lines?
Electric potential decreases in the direction of the electric field.
Electric potential increases in the opposite direction.
7. How is work related to electric potential?
Work done when moving a charge between two points is:
$W = -q\Delta V$
where $\Delta V = V_B - V_A$
8. What is electric potential difference?
Electric potential difference is the difference in electric potential between two points.
$V_{AB} = V_A - V_B$
It represents the work needed per unit charge to move a charge between the two points.
9. Can electric potential be negative?
Yes. Electric potential becomes negative near negative charges because the electric potential energy of a positive test charge decreases.
10. How do multiple charges affect electric potential?
Electric potential follows the principle of superposition:
$V = V_1 + V_2 + V_3 + \dots$
The total potential is the sum of potentials from all charges.
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