A charge Q is distributed over two concentric conducting thin spherical shells radii r and R (R > r). If the surface charge densities on the two shells are equal, the electric potential at the common centre is :

1 \over 4\pi \varepsilon _0( R + r ) \over ( R^2 + r^2 ) Q

1 \over 4\pi \varepsilon _0( R + r ) \over 2( R^2 + r^2 )Q

1 \over 4\pi \varepsilon _0( R + 2r )Q \over 2( R^2 + r^2 )

1 \over 4\pi \varepsilon _0( 2R + r ) \over ( R^2 + r^2 )Q

Electrostatics PYQ — JEE Main 2020 · 2 Sep · Shift 2 · Q45 with Solution

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Correct answer: A

Let the charges on the two shells be $q_1$ and $q_2$ .

Total charge: $q_1+q_2=Q$
Use the condition of equal surface charge densities.

Surface charge density on inner shell of radius $r$ : $\sigma_1=\frac{q_1}{4\pi r^2}$

Surface charge density on outer shell of radius $R$ : $\sigma_2=\frac{q_2}{4\pi R^2}$

Given $\sigma_1=\sigma_2$ , so $\frac{q_1}{r^2}=\frac{q_2}{R^2}$ $q_1:q_2=r^2:R^2$

Therefore, $q_1=\frac{Qr^2}{r^2+R^2},\qquad q_2=\frac{QR^2}{r^2+R^2}$
Potential at the centre due to a charged spherical shell.

For a conducting spherical shell, potential at any interior point is constant and equal to the surface potential: $V=\frac{1}{4\pi\varepsilon_0}\frac{q}{a}$ where $a$ is the radius of the shell.

So, potential at the common centre due to both shells is $V=\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r}+\frac{q_2}{R}\right)$
Substitute $q_1$ and $q_2$ .

$V=\frac{1}{4\pi\varepsilon_0}\left(\frac{Qr^2}{(r^2+R^2)r}+\frac{QR^2}{(r^2+R^2)R}\right)$

$V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2+R^2}(r+R)$

Hence, $V=\frac{1}{4\pi\varepsilon_0}\frac{(R+r)}{R^2+r^2}Q$
Match with the options.

This is exactly Option A.

Final Answer: $\boxed{V=\frac{1}{4\pi\varepsilon_0}\frac{(R+r)}{R^2+r^2}Q}$ So the correct option is A.