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Asked by aaryamanmodern | 06 Aug, 2023, 07:12: PM
answered-by-expert Expert Answer
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
 
begin mathsize 14px style E subscript e x t end subscript space plus space E subscript p space equals space 5 space cross times space 10 to the power of 5 space V divided by m end style  .............................(1)
 
Right side net electric field is
 
begin mathsize 14px style E subscript e x t end subscript space minus space E subscript p space equals space 3 space cross times space 10 to the power of 5 space space V divided by m end style ............................(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
 
begin mathsize 14px style E subscript p space equals space fraction numerator sigma over denominator 2 space epsilon subscript o end fraction space equals space 10 to the power of 5 space V divided by m end style
 
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.

begin mathsize 14px style fraction numerator sigma over denominator 2 epsilon subscript o end fraction space equals space fraction numerator Q over denominator A space cross times 2 epsilon subscript o end fraction space equals space 10 to the power of 5 space V divided by m end style
 
begin mathsize 14px style space Q space equals space A space cross times space 2 space epsilon subscript o space cross times 10 to the power of 5 space space space C end style  .......................... (3)
 
Force F on chrged metal plate due to external electric field Eext is

begin mathsize 14px style F space equals space Q space cross times space E subscript e x t end subscript space equals space 0.08 space N end style   ............................. (4)
 
Using eqn.(3) and using the value of Eext , we rewrite above eqn.(4) as
 
begin mathsize 14px style A space cross times space 2 space epsilon subscript o space cross times 10 to the power of 5 space cross times space 4 space asterisk times space 10 to the power of 5 space equals space 0.08 space space N end style
 
Above expression is simplified as
 
begin mathsize 14px style A space cross times space epsilon subscript o space equals space 10 to the power of negative 12 end exponent end style .........................(5)
 
if begin mathsize 14px style fraction numerator 1 over denominator 4 pi epsilon subscript o end fraction space end style= 9 × 109 , then begin mathsize 14px style epsilon subscript o space equals space fraction numerator 10 to the power of negative 9 end exponent over denominator 36 straight pi end fraction space F divided by m end style
 
By substituting εo in eqn.(5) , we get area of plate as begin mathsize 14px style A space equals space 3.6 space straight pi space cross times 10 to the power of negative 2 space end exponent space straight m squared end style
 
Answered by Thiyagarajan K | 07 Aug, 2023, 01:07: PM
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