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Question

A charge Q Is distributed over two concentric hollow spheres of radii r and $$ R ( R > r) $$ such that their surface densities are equal.
The potential at the common center is

A
$$ \frac {1}{ 2 \pi \epsilon_0} \frac {Q ( R +r) }{ ( R^2 +r^2 )} $$
B
$$ \frac {1}{ 4 \pi \epsilon_0} \frac {Q( R +r) }{ (R^2 +r^2 )} $$
C
$$ \frac { \sqrt {2} }{ \pi \epsilon_0} \frac {Q( R+ r)}{(R^2 + r^2 ) } $$
D
$$ \frac {1}{ \pi \epsilon_0} \frac {Q (R -r)}{(R^2 +r^2) } $$
Solution
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Correct option is A. $$ \frac { \sqrt {2} }{ \pi \epsilon_0} \frac {Q( R+ r)}{(R^2 + r^2 ) } $$
If $$ q_1 $$ and $$ q_2 $$ are the charges on the spheres of radii r and R respectively , then, by conservation of charge
$$ q_1 +q_2 =Q ..................(i) $$
But $$ \sigma_1 = \sigma_2 $$
$$ \frac {q_1}{ 4 \pi r^2} = \frac {q_2}{ 4 \pi R^2 } $$
$$ \frac {q_1}{q_2} = \frac {r^2}{R^2} .............(ii)$$
From Eqs (I) and (ii) , we get
$$ q_1 = \frac {Qr^2}{ r^2 +R^2} $$ and $$ q_2 = \frac {QR^2}{r^2+ R^2} $$
hence , potential at common center,
$$ V = V_1 + V_2 $$
$$ = \frac {1}{ 4 \pi \epsilon_0}[ \frac {q_1}{r} + \frac {q_2}{R} ] $$
$$ = - \frac {1}{ 4 \pi \epsilon_0} [ \frac {Qr}{(R^2 +r^2)} + \frac {QR}{ (R^2 + r^2 ) } ] $$
$$ = \frac {1}{ 4 \pi \epsilon_0} \frac {Q ( R+r) }{ (R^2 +r^2)} $$

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