Set 51 Problem number 20

Problem

Find the magnitude of the potential difference between point A and point B if to move .95 Coulombs of charge from A to B in 9 seconds   requires an average power of .87 watts.

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 .87 J/s, while charge is passing at ( .95 C)/( 9 sec) = .1055 C/s.
• Thus .87 J must correspond to .1055 C, so that the potential difference must be .87 J / ( .1055 C) = 8.246 J/C, or 8.246 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.