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  • 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 q size 12{q} {} from a large distance away to a distance of r size 12{r} {} from a point charge Q size 12{Q} {} , and noting the connection between work and potential W = q Δ V size 12{ left (W= - q?V right )} {} , it can be shown that the electric potential V size 12{V} {} of a point charge is

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

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

Electric potential V size 12{V} {} Of a point charge

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

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

The potential at infinity is chosen to be zero. Thus V size 12{V} {} for a point charge decreases with distance, whereas E size 12{E} {} for a point charge decreases with distance squared:

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

Recall that the electric potential V size 12{V} {} is a scalar and has no direction, whereas the electric field E 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 V size 12{V} {} is closely associated with energy, a scalar, whereas E size 12{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 size 12{ left ("nC" right )} {} to microcoulomb µC 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 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 / r size 12{V= ital "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. 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" "." {} } } {}


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. (See [link] .) 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.
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 . size 12{V= ital "kQ"/r} {}


Solving for 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. 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 "." {} } } {}


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, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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