Question

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

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

Answer 1

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

From the fig. it is clear that the plate is placed in an external electric field. Let the electric field due to plate be E, and $E_0$ is the external electric field.
Therefore,
$E_0+E=12V{m}^{-1}$(i)
$E_0-E=8V{m}^{-1}$(ii)
Solving Eqs. i) and (ii), $E=2V{m}^{-1}$ and $E_0=10V{m}^{-1}$
Now, electric field due to plate is $\sigma /2{\epsilon }_0=2$. Therefore, $\sigma =4{\epsilon }_0$