Set 51 Problem number 20

Problem

To move .825 Coulombs of charge from point A to point B in 5.9 seconds requires an average power of .12 watts. What is the magnitude of the potential difference between A and B?

Solution

Potential difference is measured in Volts, or Joules per Coulomb. Let's reason this directly by thinking about the meanings of the units:

• The work is being done at the rate of .12 J/s, while charge is passing at ( .825 C)/( 5.9 sec) = .1398 C/s.
• Thus .12 J must correspond to .1398 C, so that the potential difference must be .12 J / ( .1398 C) = .8583 J/C, or .8583 Volts.

Generalized Solution

When a charge q moves a distance `dx in a potential gradient `dV / `dx,

• the force on the charge is F = q * `dV / `dx,
• the work done is `dW = F * `dx = q * `dV / `dx * `dx, so
• the average rate at which work is done is P = `dW / `dt = F * `dx / `dt = q `dV / `dx * `dx / `dt = q `dV / `dt.

Given the power, the charge and the time interval, we can solve

• P = q `dV / `dt

for the potential difference `dV, obtaining

• `dV = P * `dt / q.

Note that P * `dt is the work done on the charge, so that P * `dt / q is the work per unit charge, which is the potential difference.