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Asked by aaryamanmodern | 06 Aug, 2023, 07:12: PM
Magnitude of electric field on the left side is greater tha the magnitude of electric field on the right side.

Hence direction of external electric field and that of electric field due to charges on left side metal plate are  same .

Direction of external electric field is opposite to the diretion of electric field due to charges on right side metal plate.

Let Eext is intensity of electric field and Ep is electric field intensity due to charge on metal plate.

Left side net electric field is

.............................(1)

Right side net electric field is

............................(2)

By solving eqn.(1) and (2) , we get

Eext = 4 × 105 V/m    and Ep = 1 × 105 V/m  ;

Electric field near charged plate of large area is

where σ  = Q / A is charge density, Q is total charge on plate ,  A is area of plate
and εo is permittivity of free space.

.......................... (3)

Force F on chrged metal plate due to external electric field Eext is

............................. (4)

Using eqn.(3) and using the value of Eext , we rewrite above eqn.(4) as

Above expression is simplified as

.........................(5)

if = 9 × 109 , then

By substituting εo in eqn.(5) , we get area of plate as

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