The electric field on two sides of a large charged plate is shown in figure. The charge density on the plate in SI units is given by `(epsilon_0` is the permittivity of free space in SI units).

Options

A. `2epsilon_0`
B. `4epsilon_0`
C. `10epsilon_0`
D. zero

Answer

B

Explanation

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 `E_0` is the external electric field. Therefore,

`E_0 + E = 12 Vm^(-1)`........(i)
`E_0 - E = 8Vm^(-1)`.......(ii)
Solving Eqs. (i) and (ii), E `= 2Vm^(-1)` and `E_0 = 10Vm^(-1)`
Now, electric field due to plate is `sigma//2epsilon_0` = 2. Therefore, `sigma = 4epsilon_0`.