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Asked by aaryamanmodern | 06 Aug, 2023, 07:12: PM

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 E_extis intensity of electric field and E_pis 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

E_ext = 4 × 10⁵ V/m and E_p = 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 ε_o 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 E_ext 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 E_ext , 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)

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 ε_o 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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