Question

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0\times10^{-22}\,\text{C/m}^{2}$. What is $E$: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Hint:

Outer regions of both plates: fields cancel so E = 0. Between the plates the fields add: E = sigma / epsilon_0.

Answer 1

Step 1: Set up the picture. Two parallel plates carry equal and opposite surface charge densities \(+\sigma\) and \(-\sigma\) on their inner faces, with \(\sigma = 17.0\times10^{-22}\,\text{C m}^{-2}\). This is the classic parallel-plate capacitor field.
Step 2: A single infinite charged sheet produces a field \(\sigma/2\epsilon_0\) on each side. Add the two sheets:
In the two outer regions the two fields point opposite ways and cancel.
In the region between the plates they point the same way and add.
Step 3 (parts a and b): Outer region of the first plate and outer region of the second plate:
\[E = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0\]
Step 4 (part c): Between the plates the fields add:
\[E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}\]
Step 5: Substitute \(\sigma = 17.0\times10^{-22}\,\text{C m}^{-2}\) and \(\epsilon_0 = 8.85\times10^{-12}\):
\[E = \frac{17.0\times10^{-22}}{8.85\times10^{-12}}\]
Step 6: Do the arithmetic:
\[E = 1.92\times10^{-10}\,\text{N C}^{-1}\]
\[\boxed{E_a = 0,\quad E_b = 0,\quad E_c = 1.92\times10^{-10}\,\text{N C}^{-1}}\]

Answer 2

Conductor surface-field route.
Step 1: For an ideal parallel-plate arrangement, all the charge sits on the inner faces and the field is confined to the gap. By Gauss's law applied to a pillbox straddling one plate's inner surface, the field just outside a conductor surface is:
\[E = \frac{\sigma}{\epsilon_0}\]
This is the field in the gap between the plates.
Step 2 (parts a and b): Outside both plates there is no enclosed net charge for a pillbox in the outer regions (field lines start on \(+\sigma\) and end on \(-\sigma\), all inside). Hence:
\[E_{\text{outer}} = 0\]
Step 3 (part c): Substitute the gap field:
\[E = \frac{17.0\times10^{-22}}{8.85\times10^{-12}} = 1.92\times10^{-10}\,\text{N C}^{-1}\]
\[\boxed{E_a = 0,\quad E_b = 0,\quad E_c = 1.92\times10^{-10}\,\text{N C}^{-1}}\]