Question

Two large parallel metal plates are 6.0 cm apart. The magnitude of the electric field between them is 600 N/C. What work is done when one electron is moved from the positive to the negative plate? Note: the charge of an electron is 1.6 x 10-19 C.

3.36 x 10-20 J



4.98 x 10-17 J



1.6 x 10-19 J



5.76 x 10-18 J

Answer 1

Approximately
`5.76 * 10^(-18)\; {\rm J}` (by the external force that moved the electron.)

Step-by-step explanation:

Let
`q` denote the magnitude of a charge in an electric field of magnitude
`E`. The magnitude of the electrostatic force on that charge would be:

`F = q\, E`.

In the electric field in this question, the magnitude of the electric force on the electron would be:

`\begin{aligned}F &= q\, E \\ &= 1.6 * 10^(-19)\; {\rm C} * 600\; {\rm N \cdot C^(-1)} \\ &= 9.6 * 10^(-17) \; {\rm N}\end{aligned}`.

Moving the electron to the other plate would thus require an external force of
`9.6 * 10^(-17)\; {\rm N}`.

This force needs to be exerted over a distance of
`6.0\; {\rm cm}` (
`6.0 * 10^(-2)\; {\rm m}`.) The direction of the motion is the same direction as that of the external force. Thus, the work that needs to be done would be:

`\begin{aligned}W &= F\, s \\ &\approx 9.6 * 10^(-17)\; {\rm N} * 6.0 * 10^(-2)\; {\rm m} \\ &\approx 5.76 * 10^(-18)\; {\rm J}\end{aligned}`.