Question

The electric field on two sides of a large charged sheet is shown in the figure. The charge density on the sheet in $\text{SI}$ units is given by $({\epsilon }_0$ is the permittivity of free space in $\text{SI}$ units $)$

• A. $2{\epsilon }_0$
• B. Zero.
• C. $4{\epsilon }_0$
• D. $10{\epsilon }_0$

Answer 1

The correct option is C $4{\epsilon }_0$

From the figure, it is clear that the sheet is placed in an external electric field.

Let the electric field due to the sheet be $E$ and $E_0$ be the external electric field.


$\therefore E+E_0=12\text{V/m}$ and $E_0-E=8\text{V/m}$

Solving these two equations we get, $E=2\text{V/m}$

We know that, electric field due to a conducting sheet is given by $$E=\frac{\sigma }{2{\epsilon }_0}$$ Thus, $\frac{\sigma }{2{\epsilon }_0}=2$ $\Rightarrow \sigma =4{\epsilon }_0$

Hence, option (b) is the correct answer.