College Physics for AP® Courses

19.3Electrical Potential Due to a Point Charge

College Physics for AP® Courses19.3 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 $qq size 12{q} {}$ from a large distance away to a distance of $rr size 12{r} {}$ from a point charge $QQ size 12{Q} {}$, and noting the connection between work and potential $W=–qΔVW=–qΔV size 12{ left (W= - q?V right )} {}$, it can be shown that the electric potential $VV size 12{V} {}$ of a point charge is

$V = kQ r ( Point Charge ) , V = kQ r ( Point Charge ) , size 12{V= { { ital "kQ"} over {r} } $$"Point Charge"$$ ,} {}$
19.37

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 $VV size 12{V} {}$ of a Point Charge

The electric potential $VV size 12{V} {}$ of a point charge is given by

$V = kQ r ( Point Charge ) . V = kQ r ( Point Charge ) . size 12{V= { { ital "kQ"} over {r} } $$"Point Charge"$$ ,} {}$
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The potential at infinity is chosen to be zero. Thus $VV size 12{V} {}$ for a point charge decreases with distance, whereas $EE size 12{E} {}$ for a point charge decreases with distance squared:

$E = F q = kQ r 2 . E = F q = kQ r 2 . size 12{ left (E=F/q right )= ital "kQ/r" rSup { size 8{2} } } {}$
19.39

Recall that the electric potential $VV size 12{V} {}$ is a scalar and has no direction, whereas the electric field $EE size 12{E} {}$ 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 $VV size 12{V} {}$ is closely associated with energy, a scalar, whereas $EE size 12{E} {}$ 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 $nCnC size 12{ left ("nC" right )} {}$ to microcoulomb $µCµC size 12{ left (µC right )} {}$ 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?

Strategy

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 $V=kQ/rV=kQ/r size 12{V= ital "kQ"/r} {}$.

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. 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. alignl { stack { size 12{V=k { {Q} over {r} } = left (9 "." "00" times "10" rSup { size 8{9} } " N" cdot m rSup { size 8{2} } /C rSup { size 8{2} } right ) { { - 3 "." "00" times "10" rSup { size 8{ - 9} } " C"} over {5 "." "00" times "10" rSup { size 8{"–2"} } " m"} } } {} # = - "540"" V" "." {} } } {}
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Discussion

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

Figure 19.11 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 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

$V = kQ r . V = kQ r . size 12{V= ital "kQ"/r} {}$
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Solution

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. alignl { stack { size 12{Q= { { ital "rV"} over {k} } = { { left (0 "." "12"" m" right ) left ("100"´"10" rSup { size 8{3} } " V" right )} over {9 "." "00"´"10" rSup { size 8{9} } " N" cdot m rSup { size 8{2} } /C rSup { size 8{2} } } } } {} # =1 "." "39"´"10" rSup { size 8{-6} } " C=1" "." "39 "mC "." {} } } {}
19.42

Discussion

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 voltages in both of these examples could be measured with a meter that compares the measured potential with ground potential. Ground potential is often taken to be zero (instead of taking the potential at infinity to be zero). It is the potential difference between two points that is of importance, and very often there is a tacit assumption that some reference point, such as Earth or a very distant point, is at zero potential. As noted in Electric Potential Energy: Potential Difference, this is analogous to taking sea level as $h=0h=0 size 12{h=0} {}$ when considering gravitational potential energy, $PEg=mghPEg=mgh size 12{"PE" rSub { size 8{g} } = ital "mgh"} {}$.