Electric Potential Due to Point Charges

4.4 The Electric Potential

4.4.1 Electric Potential Due to Point Charges

For a chargeQlocated at the origin, the electric field intensity anywhere in space is given byEq. (4.1). The resulting electric field is shown inFigure 4.9. Now, consider two pointsa andb at distancesR2and R1, respectively, from the origin.

To calculate the potential, we choosepath(1) inFigure4.9and integrate from pointato pointcand then from pointcto pointb. We get

Vba¼VbVa¼ðVcVaÞ þðVbVcÞ ¼ ðc

a

E

^dl1ðb

c

E

^{dl2 ½ V ð4:21Þ}

In this case, calculation is best carried out in spherical coordinates. Withdl¼R^dRþθ^Rdθþϕ^Rsinθdϕ,dl1anddl2are

dl1¼R^dR, dl2¼θ^Rdθ ð4:22Þ

Before performing the integration inEq. (4.21), we must evaluate the integrand. In the first integral inEq. (4.21),Eand dl1are in the same direction; therefore,E

^{dl1¼QdR/4πε0|R2}|. In the second integral,Eis in the positiveRdirection anddl2

is in the positiveθ direction. Thus, E

^{dl2¼0 sinceE anddl2}are perpendicular to each other. Substituting these into Eq. (4.21), we get

Vba¼ ðR_e

R_a

QdR 4πε0

R² ½ V ð4:23Þ

Noting thatRc¼Rb(pointsbandcare at same distance from the origin) and |R|¼R, we get

Vba ¼ ðRb

R_a

Q

4πε0R²dR¼ Q 4πε0Rb

Q 4πε0Ra

0

@

1 A¼ Q

4πε0

1 Rb

1 Ra

0

@

1

A¼VbVa ½ V ð4:24Þ

First, we should note that sinceRb<Ra, this potential is positive. In other words, the closer to the point charge we get, the largerVbbecomes. In fact, atRb!0, the potentialVbtends to infinity.

So far, we dealt strictly with potential differences, although inEq. (4.24),VbandVaare the absolute potentials at pointsa andb, respectively. In general, we cannot calculate these potentials independently. However, in the case of a point charge and other finite-size charge distributions, it is possible to calculate the potential difference between pointband pointa, with pointaat1. In this case,Va¼0 and we get the absolute potential at pointbas

Vb¼ ðRb

1

Q

4πε0R²dR¼ Q 4πε0Rb

½ V ð4:25Þ

The potential at pointb is now the work per unit charge required to bring a unit charge (test charge) from infinity to pointb. Therefore, although we call this an absolute potential, it is a potential difference between pointband the reference point at infinity.

Calculation of absolute potentials is often possible and sometimes preferred because once absolute potentials are known, potential differences are easy to calculate. This is also compatible with our normal practice of assigning a reference value (such as ground potential in a circuit) and measuring all potentials with respect to the reference value. However, there will be instances where the zero potential reference at1is either impossible to satisfy or inconvenient to apply. For example, an infinitely long line or infinite plane extends to infinity. Now, the potential at infinity cannot be assumed to be zero. But the reference point does not have to be at infinity and does not have to be zero. For example, if you measure potential in an electric circuit (voltage), the measurement is with respect to the“ground”or“chassis”of the instrument. We usually assume ground is at zero potential but not necessarily so. The chassis itself may be at a certain potential above ground level and planet Earth may be at yet another potential level with respect to the stars. Other reference values and other locations in space can then be chosen for this purpose, as we shall see in examples.

154 4 Gauss’s Law and the Electric Potential

All relations given so far were defined with the charge at the origin. This is obviously only a convenient choice. If the charge is located at any other point in space, the use of position vectors allows a more general expression. Referring to Figure4.10, we can write for the chargeQ,

Vb¼ Q

4πε0

R_bR⁰ and Va¼ Q 4πε0

R_aR⁰ ½ V ð4:26Þ

and the potential difference between the two points is VbVa¼ Q

4πε0

1 RbR⁰

1

RaR⁰

!

½ V ð4:27Þ

Perhaps the most remarkable aspect of the results obtained above is that the potential difference is independent of the path we take to evaluate it. Two paths are shown inFigure4.9. On either path shown, the scalar product betweenEanddlcan be separated into a component along the electric field intensity and one perpendicular to the electric field intensity and only the component along the electric field intensity contributes to the scalar product. Thus, movement in the radial direction changes the potential (contributes to the integral), whereas movement of the test charge perpendicular to the radial direction (along any circular path of constant radius) does not change the potential. The latter aspect of the field has two consequences:

(1) The potential only depends on the radial distances between pointsaandband the point charge, regardless of path taken.

This is true in any static electric field, regardless of how it may have been generated.

(2) The potential on any spherical surface enclosing a point charge (centered at the charge) is constant. These spheres are surfaces of constant potential. We note that potential is constant on lines or surfaces perpendicular to the electric field intensity. In other configurations, the lines of constant potential may not be spherical, but they are always perpendicular to the electric field intensity.

Now, for the first advantage in using the potential: its scalar nature. Consider first a number of point charges as shown in Figure 4.11. The potential at a point P(R) in space is a scalar summation of the potentials of the individual charges.

Extending the result of the single point charge, we can write Vð Þ ¼R Q₁

4πε0

RR1

þ Q2

4πε0

RR2

þ Q3

4πε0

RR3

þ þ Qn

4πε0

RRn

½ V ð4:28Þ

or, in a more compact form,

Vð Þ ¼R 1 4πε0

Xⁿ

i¼1

Q_i RRi

j j ½ V ð4:29Þ

where the distance |R–R_i| is simply the distance between theith charge and pointPas shown inFigure4.11. The potential of each charge is calculated as if all other charges do not exist and the principle of superposition is used to find the total potential.

Q

x y

z

a b

Ra

Rb

R^'

R^b−R^'

Ra−R^'

Figure 4.10 Use of position vectors to calculate the potentials at pointsaandb

Example 4.7Potential of Point Charges Three point charges are arranged as shown inFigure4.12. Calculate the potential at a general pointP(x,y) in the plane of the charges.

Solution: The potential due to each point charge is calculated separately and summed up to obtain the total potential due to the three charges.

The potential at a distanceRfrom a point charge is V¼ Q

4πε0R ½ V

whereRis the distance between the point charge and the location at which the potential is calculated. FromFigure4.12, we have

RA¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xþa ð Þ²þy², q

RB¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xa ð Þ²þy² q

, Rc¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þðybÞ² q

½ m The potentials are

VPA¼ 2Q

4πε0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xþa ð Þ²þy²

q , VPB¼ 3Q

4πε0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xa ð Þ²þy²

q , VPC ¼ Q

4πε0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þyb²

q ½ V

The potential atPis

VP¼VPAþVPBþVPC ¼ Q

4πε0 2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xþa ð Þ²þy²

q 3

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xa ð Þ²þy²

q þ 1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þðybÞ² q

2 64

3 75 ½ V P

R x y

z

R1 R2

R−R₁ R−R₂

R−Rⁿ Rn

Q_n Q₁

Q2

Figure 4.11 Calculation of potential at pointPdue tonpoint charges

Q

− −

y

x P(x,y)

A

B C

R_A

R_B R_C

b a a

2

3Q Q

Figure 4.12 Calculation of potential at pointP(x,y) due to the three charges shown

156 4 Gauss’s Law and the Electric Potential

This simple calculation indicates the ease with which the potential due to any number of point charges may be calculated.

Note also that in the final result, the term in the square brackets is a geometrical term and does not include charge or material properties. This will be exploited inChapter6to find numerical solutions to a number of much more complex problems.

Exercise 4.3 Three charges, each equal toQ¼10^–7C, are located on thexaxis atx¼0,x¼1 m, andx¼2 m, as shown inFigure4.13. Calculate the potential at (x¼0, y¼1).

Answer 1,936 V.