A large charged metal sheet is placed in a uniform electric field, perpendicular to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet is `E_1 = 5 xx 10^5 Vm^(-1)` and on the right it is `E_2 = 3 xx 10^5 V m^(-1)`. The sheet experiences a net electric force of 0.08 N. Find the area of one face of the sheet. Assume the external field to remain constant after introducing the large sheet. Use
`(1/(4piepsilon_0)) = 9 xx 10^(9) Nm^(2)C^(-2)`

Options

A. `3.6 pi xx 10^(-2)m^2`
B. `0.9 pi xx 10^(-2)m^2`
C. `1.8 pi xx 10^(-2)m^2`
D. none

Answer

A

Explanation

a. Let external field be `E_0` and surface charge density of sheet
be `sigma` (including both surfaces). So
`E_(P) = sigma/(2epsilon_0)`

Electric field due to sheet
`E_1 = E_0 - E_p`.......(i)
`-E_2 = E_0 + E_(P)`........(ii)
From (i) and (ii),
`E_0 = (E_1 - E_2)/2 = 10^5 Vm^(-1)`
and `E_(P) = (-E_1-E_2)/2 = -4 xx 10^5 Vm^(-1)`
`sigma/(2epsilon_0) = - 4 xx 10^5 or sigma = - 8 epsilon_0 xx 10^5`
force on sheet
`F = qE_0` or `0.08 = sigma AE_0`
or `0.08 = 8epsilon_0 xx 10^5 A xx 10^5`
or `A = (10^(-12))/epsilon_0 = 10^(-12) xx 36 xx 10^9 pi = 3.6 pi xx 10^(-2) m^(2)`.