Question

Two large metal sheets having surface charge density +2$\sigma$ and -2$\sigma$ are kept parallel to each other at a small separation distance d. The electric field at any point in the region between the plates is

• A. $\sigma$/$\epsilon_0$
• B. 2$\sigma$/$\epsilon_0$
• C. $\sigma$/d
• D. $\sigma$/2d

Answer 1

Answer: (B)

2$\sigma$/$\epsilon_0$

Answer 2

To find the electric field between two large parallel metal sheets, we can use the principle of superposition.

The electric field due to a single infinite plane sheet with uniform surface charge density $\Sigma$ is given by:

$E = \frac{|\Sigma|}{2\epsilon_0}$

The direction of the electric field is perpendicular to the sheet, pointing away from a positively charged sheet and towards a negatively charged sheet.

Let the first metal sheet have a surface charge density $\Sigma_1 = +2\sigma$. The electric field ($E_1$) due to this sheet at any point in the region between the plates will point away from it.

$E_1 = \frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$

Let the second metal sheet have a surface charge density $\Sigma_2 = -2\sigma$. The electric field ($E_2$) due to this sheet at any point in the region between the plates will point towards it.

$E_2 = \frac{|-2\sigma|}{2\epsilon_0} = \frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$

Since both electric fields $E_1$ and $E_2$ point in the same direction in the region between the plates, the net electric field ($E_{net}$) is their sum:

$E_{net} = E_1 + E_2 = \frac{\sigma}{\epsilon_0} + \frac{\sigma}{\epsilon_0} = \frac{2\sigma}{\epsilon_0}$

The small separation distance 'd' does not affect the magnitude of the electric field for large sheets, as the field is uniform between them.