Set 51 Problem number 16

Problem

What is the force per Coulomb acting on a charge of 30 Coulombs in the presence of a 84000 V/m potential gradient?

What symbolic expression give us the force per Coulomb on a charge q in the presence of a potential gradient dV/dx?

Show that a volt/meter is identical to a Newton/Coulomb.

Solution

The force on the 30 Coulomb charge (see preceding problem) is F = q (`dV/`ds) = 30 C ( 84000 V/m) = 2520000 C V/m = 2520000 C (J/C) / m = 2520000 J/m = 2520000 N m / m = 2520000 N.

The force per Coulomb is 2520000 N / ( 30 C) = 84000 N / C.

The force on a charge q in the presence of a potential gradient dV/dx is F = q (dV/dx), so the force per unit charge is

F/q = q (dV/dx) / q = dV/dx.

F / q clearly has units of Newtons / Coulomb; dV/dx has units of Volts / meter.

We note the unit calculation

units: V / m = (J/C) / m = (N m / C) / m = N / C.

Generalized Solution

As found in the preceding problem, the force on a charge of q Coulombs in a potential gradient dV/dx is (q Coulombs) * dV/dx.

The force per Coulomb is found by dividing by the number q of Coulombs, which gives force per Coulomb dV / dx.

Thus dV / dx, which measures volts per meter, is also a measure of Newtons per Coulomb.

Since a volt is a Joule per Coulomb, a volt per meter turns out to be identical to a Newton per Coulomb.

Explanation in terms of Figure(s), Extension

The figure below shows a charge q moving through a displacement `ds in the direction of a potential gradient dV / dx.

The potential difference is the product of the potential gradient and the displacement: `dV = dV / dx * `ds.
The work done on the charge is therefore `dW = q `dV; this work is also the product `dW = Fave * `ds of the displacement and the average force on the charge.
Setting these two expressions for work `dW equal we arrive at the conclusion that Fave = q dV/dx = q E.

E = dV / dx is called the electric field. Since E = Fave / q, we see that the electric field E is the force per unit charge experienced by a charge.