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RLC Circuit

Two large, pa...

Question

Two large, parallel conducting plates X and Y, kept close to each other, are given charges $Q_{1}$ and $Q_{2} (Q_{1} > Q_{2})$ . The four surfaces of the plates are A, B, C and D, as shown in figure. Then :

The charge on A is

\frac{1}{2} (Q_{1} + Q_{2})

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The charge on B is

\frac{1}{2} (Q_{1} - Q_{2})

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The charge on C is -

\frac{1}{2} (Q_{1} - Q_{2})

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The charge on D is

\frac{1}{2} (Q_{1} + Q_{2})

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Solution

The correct options are
A The charge on B is $\frac{1}{2} (Q_{1} - Q_{2})$ .
B The charge on A is $\frac{1}{2} (Q_{1} + Q_{2})$ .
C The charge on C is - $\frac{1}{2} (Q_{1} - Q_{2})$ .
D The charge on D is $\frac{1}{2} (Q_{1} + Q_{2})$
Let charge for surface A $= Q_{1} - q$ , for surface B $= q$ , for surface C $= Q_{2} - q^{'}$ and for surface D $= q^{'}$ .
As the field inside X plate is zero, $\frac{1}{2 A ϵ_{0}} (Q_{1} - q - q - Q_{2} + q^{'} - q^{'}) = 0$
$- 2 q = Q_{2} - Q_{1} \Rightarrow q = \frac{1}{2} (Q_{1} - Q_{2})$
similarly for Y plate, $\frac{1}{2 A ϵ_{0}} (q^{'} - Q_{2} + q^{'} - q - Q_{1} + q) = 0$
or $2 q^{'} = Q_{1} + Q_{2} \Rightarrow q^{'} = \frac{1}{2} (Q_{1} + Q_{2})$
Thus, charge on A is $Q_{1} - q = Q_{1} - \frac{1}{2} (Q_{1} - Q_{2}) = \frac{1}{2} (Q_{1} + Q_{2})$ .
charge on B is $q = \frac{1}{2} (Q_{1} - Q_{2})$
charge on C is $Q_{2} - q^{'} = Q_{2} - \frac{1}{2} (Q_{1} + Q_{2}) = - \frac{1}{2} (Q_{1} - Q_{2})$
Charge on D is $q^{'} = \frac{1}{2} (Q_{1} + Q_{2})$

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