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By the end of this section, you will be able to:
  • Calculate the potential due to a point charge
  • Calculate the potential of a system of multiple point charges
  • Describe an electric dipole
  • Define dipole moment
  • Calculate the potential of a continuous charge distribution

Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (such as charge 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.

We can use calculus to find the work needed to move a test charge q from a large distance away to a distance of r from a point charge q . Noting the connection between work and potential W = q Δ V , as in the last section, we can obtain the following result.

Electric potential V Of a point charge

The electric potential V of a point charge is given by

V = k q r ( point charge )

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

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

E = F q t = k q r 2 .

Recall that the electric potential V is a scalar and has no direction, whereas the electric field E 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 V is closely associated with energy, a scalar, whereas E is closely associated with force, a vector.

What voltage is produced by a small charge on a metal sphere?

Charges in static electricity are typically in the nanocoulomb (nC) to microcoulomb ( μ C ) range. What is the voltage 5.00 cm away from the center of a 1-cm-diameter solid metal sphere that has a –3.00-nC static charge?

Strategy

As we discussed in Electric Charges and Fields , 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 = k q r .

Solution

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 .

Significance

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.

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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 ( [link] ). What excess charge resides on the sphere? (Assume that each numerical value here is shown with three significant figures.)

The figure shows the parts of Van de Graaff generator.
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.

Strategy

The potential on the surface is 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 = k q r .

Solution

Solving for q and entering known values gives

q = r V k = ( 0.125 m ) ( 100 × 10 3 V ) 8.99 × 10 9 N · m 2 /C 2 = 1.39 × 10 −6 C = 1.39 μ C .

Significance

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.

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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