Question

The electric field intensity on two sides of an uniformly charged infinite conducting sheet is as shown in figure. The surface charge density of sheet is

• A. $13 \epsilon_0 \frac{C}{m^2}$
• B. $2 \epsilon_0 \frac{C}{m^2}$
• C. $4 \epsilon_0 \frac{C}{m^2}$

Answer 1

Answer: (C)

Option C $4\epsilon_0 \frac{C}{m^2}$

Answer 2

For an isolated conducting sheet in an external field, the net electric fields just outside on the two sides can be written as

$$E_{\text{left}} = E_0 + \frac{\sigma}{2\epsilon_0} \quad \text{and} \quad E_{\text{right}} = E_0 - \frac{\sigma}{2\epsilon_0},$$

where $E_0$ is the external field. The conductor is an equipotential so the difference in the net fields is solely due to the discontinuity caused by the surface charge. Thus, subtracting the two equations we have

$$E_{\text{right}} - E_{\text{left}} = -\frac{\sigma}{\epsilon_0}.$$

Taking magnitudes (and noting that the directions of the fields are oppositely directed),

$$|E_{\text{right}} - E_{\text{left}}| = \frac{\sigma}{\epsilon_0}.$$

Given that the field magnitudes are 15 V/m and 11 V/m, we get

$$15 - 11 = \frac{\sigma}{\epsilon_0} \quad \Longrightarrow \quad 4 = \frac{\sigma}{\epsilon_0}.$$

Therefore,

$$\sigma = 4\,\epsilon_0 \quad \left(\frac{C}{m^2}\right).$$

Thus, the correct option is C) $4\epsilon_0 \frac{C}{m^2}$.