Set 51 Problem number 14

Problem

What is the average magnitude of the electric field between points A and B, if to move a charge of -.7208 C parallel to the field, without dissipation of energy, from A to B, a distance of 23.5 m, requires -35.64 Joules of energy?

Solution

The average strength of an electric field is measured as the average force per unit of charge.

The average force associated with -35.64 Joules of work over a displacement of 23.5 meters is (-35.64 J)/( 23.5 m) = (-35.64 N m)/( 23.5 m) = -1.518 N.
The magnitude of the field strength associated with a force of -1.518 N on a charge of -.7208 C is |(-1.518 N)/(-.7208 C)| = 2.104 N/C.

Generalized Solution

If a charge Q moves between two points and `dW Joules of work are done as the charge moves through a distance `ds, then we easily determine the average force F = `dW / `ds on the charge. The electric field is the force per unit charge (measured in Newtons/Coulomb), or in this case

E = F / Q = (`dW / `ds) / Q = `dW / (Q `ds).

Note that since `dW / Q = `dV, the potential difference, we have

E = `dW / (Q `ds) = `dV / `ds.

This shows that the electric field can also be expressed in units of volts per unit distance (i.e., volts / meter).

Explanation in terms of Figure(s), Extension

The figure below shows a charge Q moving from a potential V1 to a higher potential V2. The potential difference is `dV = V2 - V1, and the work required to move the charge is `dW = Q `dV (`dV is measured in Joules of work required per Coulomb of charge, so work must be the product of Q (Coulombs) and `dV (Joules / Coulomb) ).

If the distance `ds between points is known, we can calculate the force F = `dW / `ds. Then from the force F and the charge Q we can find the electric field strength E = | F | / | Q |.

This electric field is also seen to have magnitude E = | `dV / Q |.