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 ( $ε_{0}$ is the permittivity of free space in S.I. units)

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 ( ε 0 is the permittivity of free space in S.I. units)

TestHub · Accepted Answer

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 E₀is the external electric field.

$∴ E_{0} + E = 12 {V m}^{- 1}$

$E_{0} - E = 8 {V m}^{- 1}$

Solving equations, E = 2 V m^-1and E₀= 10 V m^-1

Now, electric field due to plate $= \frac{σ}{2 ε_{0}} = 2$

$∴ σ = 4 ε_{0}$

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