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College Physics 2e

19.3 Electrical Potential Due to a Point Charge

College Physics 2e19.3 Electrical Potential Due to a Point Charge

Learning Objectives

By the end of this section, you will be able to:

  • Explain point charges and express the equation for electric potential of a point charge.
  • Distinguish between electric potential and electric field.
  • Determine the electric potential of a point charge given charge and distance.

Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (like on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. Using calculus to find the work needed to move a test charge qq from a large distance away to a distance of rr from a point charge QQ, and noting the connection between work and potential W=qΔVW=qΔV, it can be shown that the electric potential VV of a point charge is

V = kQ r ( Point Charge ) , V = kQ r ( Point Charge ) ,

where k is a constant equal to 9.0 × 10 9 N · m 2 / C 2 9.0 × 10 9 N · m 2 / C 2 .

Electric Potential V V of a Point Charge

The electric potential VV of a point charge is given by

V = kQ r ( Point Charge ) . V = kQ r ( Point Charge ) .

The potential at infinity is chosen to be zero. Thus VV for a point charge decreases with distance, whereas EE for a point charge decreases with distance squared:

E = F q = kQ r 2 . E = F q = kQ r 2 .

Recall that the electric potential VV is a scalar and has no direction, whereas the electric field EE is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact that VV is closely associated with energy, a scalar, whereas EE is closely associated with force, a vector.

Example 19.6

What Voltage Is Produced by a Small Charge on a Metal Sphere?

Charges in static electricity are typically in the nanocoulomb nCnC to microcoulomb µCµC range. What is the voltage 5.00 cm away from the center of a 1-cm diameter metal sphere that has a −3.00 nC −3.00 nC static charge?


As we have discussed in Electric Charge and Electric Field, charge on a metal sphere spreads out uniformly and produces a field like that of a point charge located at its center. Thus we can find the voltage using the equation V=kQ/rV=kQ/r.


Entering known values into the expression for the potential of a point charge, we obtain

V = k Q r = 8.99 × 10 9 N · m 2 / C 2 –3.00 × 10 –9 C 5.00 × 10 –2 m = –539 V. V = k Q r = 8.99 × 10 9 N · m 2 / C 2 –3.00 × 10 –9 C 5.00 × 10 –2 m = –539 V.


The negative value for voltage means a positive charge would be attracted from a larger distance, since the potential is lower (more negative) than at larger distances. Conversely, a negative charge would be repelled, as expected.

Example 19.7

What Is the Excess Charge on a Van de Graaff Generator

A demonstration Van de Graaff generator has a 25.0 cm diameter metal sphere that produces a voltage of 100 kV near its surface. (See Figure 19.7.) What excess charge resides on the sphere? (Assume that each numerical value here is shown with three significant figures.)

The figure shows a Van de Graaff generator. The generator consists of a flat belt running over two metal pulleys. One pulley is positioned at the top and another at the bottom. The upper pulley is surrounded by an aluminum sphere. The aluminum sphere has a diameter of twenty five centimeters. Inside the sphere, the upper pulley is connected to a conductor which in turn is connected to a voltmeter for measuring the potential on the sphere. The lower pulley is connected to a motor. When the motor is switched on, the lower pulley begins turning the flat belt. The Van de Graaff generator with the above described setup produces a voltage of one hundred kilovolts. The potential on the surface of the sphere will be the same as that of a point charge at the center which is twelve point five centimeters away from the center. Thus the excess charge is calculated using the formula Q equals r times V divided by k.
Figure 19.7 The voltage of this demonstration Van de Graaff generator is measured between the charged sphere and ground. Earth’s potential is taken to be zero as a reference. The potential of the charged conducting sphere is the same as that of an equal point charge at its center.


The potential on the surface will be the same as that of a point charge at the center of the sphere, 12.5 cm away. (The radius of the sphere is 12.5 cm.) We can thus determine the excess charge using the equation

V = kQ r . V = kQ r .


Solving for Q Q and entering known values gives

Q = rV k = 0 . 125 m 100 × 10 3 V 8.99 × 10 9 N · m 2 / C 2 = 1.39 × 10 –6 C=1.39 µC. Q = rV k = 0 . 125 m 100 × 10 3 V 8.99 × 10 9 N · m 2 / C 2 = 1.39 × 10 –6 C=1.39 µC.


This is a relatively small charge, but it produces a rather large voltage. We have another indication here that it is difficult to store isolated charges.

The reference V = 0 in both of these examples was set at infinity. If these voltages were to be measured, this would be done with a meter that compares the measured potential with a reference that is not at infinity, such as ground potential. It is the potential difference between two points that is of importance, and very often there is a tacit assumption as to that reference point, such as Earth or a very distant point. As noted in Electric Potential Energy: Potential Difference, this is analogous to taking sea level as h=0h=0 when considering gravitational potential energy, PEg=mghPEg=mgh.

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