JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 2)
A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be -
$${{Q\left( {{a^2} + {b^2} + {c^2}} \right)} \over {4\pi {\varepsilon _0}\left( {{a^3} + {b^3} + {c^3}} \right)}}$$
$${Q \over {4\pi {\varepsilon _0}\left( {a + b + c} \right)}}$$
$${Q \over {12\pi {\varepsilon _0}}}{{ab + bc + ca} \over {abc}}$$
$${{Q\left( {a + b + c} \right)} \over {4\pi {\varepsilon _0}\left( {{a^2} + {b^2} + {c^2}} \right)}}$$
Explanation
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Potential at point P, V = $${{k{Q_a}} \over a} + {{k{Q_b}} \over b} + {{k{Q_c}} \over c}$$
$$ \because $$ Qa : Qb : Qc : : a2 : b2 : c2
{since $$\sigma $$a = $$\sigma $$b = $$\sigma $$c}
$$ \therefore $$ Qa = $$\left[ {{{{a^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]$$Q
Qb = $$\left[ {{{{b^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]$$ Q
Qc = $$\left[ {{{{c^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]$$ Q
V = $${Q \over {4\pi { \in _0}}}\left[ {{{\left( {a + b + c} \right)} \over {{a^2} + {b^2} + {c^2}}}} \right]$$
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