Question 426: Electrostatics & Capacitance

Question

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 :

Accepted Answer

Let the charges on inner and outer spheres are Q
1
and Q
2
.
$$ 	herefore $$ Q
1
+ Q
2
= Q  ...(1)
Since charge density '$$\sigma $$' is same for both spheres, so
$$\sigma $$ = $${{{Q_1}} \over {4\pi {r^2}}} = {{{Q_2}} \over {4\pi {R^2}}}$$
$$ \Rightarrow $$ $${{{Q_1}} \over {{Q_2}}} = {{{r^2}} \over {{R^2}}}$$ ....(2)
From (1) and (2),
$${{{Q_2}{r^2}} \over {{R^2}}} + {Q_2} = Q$$
$$ \Rightarrow $$ Q
2
= $${{Q{R^2}} \over {\left( {{r^2} + {R^2}} ight)}}$$
$$ 	herefore $$ Q
1
= $${{{r^2}} \over {{R^2}}}.{{Q{R^2}} \over {\left( {{r^2} + {R^2}} ight)}}$$ = $${{Q{r^2}} \over {\left( {{r^2} + {R^2}} ight)}}$$
Potential at centre 'O' = $${{k{Q_1}} \over r} + {{k{Q_2}} \over R}$$
= $$k\left[ {{{Q{r^2}} \over {r\left( {{r^2} + {R^2}} ight)}} + {{Q{R^2}} \over {R\left( {{r^2} + {R^2}} ight)}}} ight]$$
= $${{{kQ\left( {r + R} ight)} \over {\left( {{r^2} + {R^2}} ight)}}}$$
= $${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} ight)} \over {\left( {{R^2} + {r^2}} ight)}}$$Q

Answer

$${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} \right)} \over {\left( {{R^2} + {r^2}} \right)}}$$Q

Answer

$${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} \right)} \over {2\left( {{R^2} + {r^2}} \right)}}Q$$

Answer

$${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + 2r} \right)Q} \over {2\left( {{R^2} + {r^2}} \right)}}$$

Answer

$${1 \over {4\pi {\varepsilon _0}}}{{\left( {2R + r} \right)} \over {\left( {{R^2} + {r^2}} \right)}}Q$$

Electrostatics & Capacitance

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 :

Choose the correct answer: