JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A charge \(Q\) is distributed over three concentric spherical shell 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
- A \(\frac{Q}{{12\pi \,{ \in _0}}}\frac{{ab + bc + ca}}{{abc}}\)
- B \(\frac{{Q\,\left( {{a^2} + {b^2} + {c^2}} \right)}}{{4\pi \,{ \in _0}\,\left( {{a^3} + {b^3} + {c^3}} \right)\,}}\)
- C \(\frac{Q}{{4\pi \,{ \in _0}\,\left( {a + b + c} \right)\,}}\)
- D \(\frac{{Q\,\left( {a + b + c} \right)}}{{4\pi \,{ \in _0}\,\left( {{a^2} + {b^2} + {c^2}} \right)\,\,}}\)
Answer & Solution
Correct Answer
(D) \(\frac{{Q\,\left( {a + b + c} \right)}}{{4\pi \,{ \in _0}\,\left( {{a^2} + {b^2} + {c^2}} \right)\,\,}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Q}_{1}+\mathrm{q}_{2}+\mathrm{Q}_{3}=\mathrm{Q}.........(1)\) \(\frac{\mathrm{Q}_{1}}{4 \pi \mathrm{a}^{2}}=\frac{\mathrm{Q}_{2}}{4 \pi \mathrm{b}^{2}}=\frac{\mathrm{Q}_{3}}{4 \pi \mathrm{c}^{2}}=\mathrm{k}.........(2)\) Subs. \(Q_{1}, Q_{2}, Q_{3}\) in \(( 1)\)…
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