Question

The electric field on two sides of a large charged plate is shown in the diagram. The charge density on the plate in S.I. units is given by ( $\epsilon _{0}$ is the permittivity of free space in S.I. units)

• A. $2 ε_{0}$
• B. $4 \epsilon _{0}$
• C. $10 ε_{0}$
• D. zero

Answer 1

Answer: (B)

From the figure, it is clear that the plate is placed in an external electric field. Let the electric field due to plate be E, and E₀ is the external electric field.



$∴ \, \, \, \text{E}_{0} + \text{E} = 1 2 V m^{- 1}$
$\text{E}_{0} - \text{E} = 8 V m^{- 1}$
Solving equations, E = 2 V m^-1 and E₀ = 10 V m^-1
Now, electric field due to plate $= \frac{\sigma }{2 \epsilon _{0}} = 2$
$∴ \, \sigma = 4 \epsilon _{0}$