JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 3)
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 :
_2nd_September_Evening_Slot_en_3_1.png)
$${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} \right)} \over {\left( {{R^2} + {r^2}} \right)}}$$Q
$${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} \right)} \over {2\left( {{R^2} + {r^2}} \right)}}Q$$
$${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + 2r} \right)Q} \over {2\left( {{R^2} + {r^2}} \right)}}$$
$${1 \over {4\pi {\varepsilon _0}}}{{\left( {2R + r} \right)} \over {\left( {{R^2} + {r^2}} \right)}}Q$$
Explanation
Let the charges on inner and outer spheres are Q1 and Q2.
$$ \therefore $$ Q1 + Q2 = 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),
$${{{Q_2}{r^2}} \over {{R^2}}} + {Q_2} = Q$$
$$ \Rightarrow $$ Q2 = $${{Q{R^2}} \over {\left( {{r^2} + {R^2}} \right)}}$$
$$ \therefore $$ Q1 = $${{{r^2}} \over {{R^2}}}.{{Q{R^2}} \over {\left( {{r^2} + {R^2}} \right)}}$$ = $${{Q{r^2}} \over {\left( {{r^2} + {R^2}} \right)}}$$
Potential at centre 'O' = $${{k{Q_1}} \over r} + {{k{Q_2}} \over R}$$
= $$k\left[ {{{Q{r^2}} \over {r\left( {{r^2} + {R^2}} \right)}} + {{Q{R^2}} \over {R\left( {{r^2} + {R^2}} \right)}}} \right]$$
= $${{{kQ\left( {r + R} \right)} \over {\left( {{r^2} + {R^2}} \right)}}}$$
= $${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} \right)} \over {\left( {{R^2} + {r^2}} \right)}}$$Q
$$ \therefore $$ Q1 + Q2 = 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),
$${{{Q_2}{r^2}} \over {{R^2}}} + {Q_2} = Q$$
$$ \Rightarrow $$ Q2 = $${{Q{R^2}} \over {\left( {{r^2} + {R^2}} \right)}}$$
$$ \therefore $$ Q1 = $${{{r^2}} \over {{R^2}}}.{{Q{R^2}} \over {\left( {{r^2} + {R^2}} \right)}}$$ = $${{Q{r^2}} \over {\left( {{r^2} + {R^2}} \right)}}$$
Potential at centre 'O' = $${{k{Q_1}} \over r} + {{k{Q_2}} \over R}$$
= $$k\left[ {{{Q{r^2}} \over {r\left( {{r^2} + {R^2}} \right)}} + {{Q{R^2}} \over {R\left( {{r^2} + {R^2}} \right)}}} \right]$$
= $${{{kQ\left( {r + R} \right)} \over {\left( {{r^2} + {R^2}} \right)}}}$$
= $${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} \right)} \over {\left( {{R^2} + {r^2}} \right)}}$$Q
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