Question:
A point charge of magnitude \(+1\,\mu\mathrm{C}\) is fixed at \((0,0,0)\). An isolated uncharged spherical conductor, is fixed with its center at \((4,0,0)\). The potential and the induced electric field at the centre of the sphere is :
A point charge of magnitude \(+1\,\mu\mathrm{C}\) is fixed at \((0,0,0)\). An isolated uncharged spherical conductor, is fixed with its center at \((4,0,0)\). The potential and the induced electric field at the centre of the sphere is :
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The net field inside a conductor is zero, so the induced field cancels the point charge's field. The induced charges sum to zero, so they contribute zero potential at the centre.
Updated On: Oct 31, 2026
- \(1.8\times10^{5}\) V and \(-5.625\times10^{6}\) V/m
- 0 V and 0 V/m
- \(2.25\times10^{5}\) V and \(-5.625\times10^{6}\) V/m
- \(2.25\times10^{5}\) V and 0 V/m
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The Correct Option is C
Solution and Explanation
The distance of the charge from the sphere's centre is 4 (units not printed; the options correspond to 4 cm \(=0.04\) m).
Potential at the centre: the induced charges lie on the surface at the same distance \(R\) from the centre and sum to zero (the sphere is uncharged), so their potential at the centre is \(\frac{k\cdot0}{R}=0\). Hence \(V=\frac{kq}{r}=\frac{9\times10^9\times10^{-6}}{0.04}=2.25\times10^5\) V.
Field of the point charge at the centre: \(E=\frac{kq}{r^2}=\frac{9\times10^9\times10^{-6}}{(0.04)^2}=5.625\times10^6\) V/m, directed along +x.
The net field inside a conductor is zero, so the induced field is \(-5.625\times10^6\) V/m.
Answer: (3) \(2.25\times10^{5}\) V and \(-5.625\times10^{6}\) V/m
Potential at the centre: the induced charges lie on the surface at the same distance \(R\) from the centre and sum to zero (the sphere is uncharged), so their potential at the centre is \(\frac{k\cdot0}{R}=0\). Hence \(V=\frac{kq}{r}=\frac{9\times10^9\times10^{-6}}{0.04}=2.25\times10^5\) V.
Field of the point charge at the centre: \(E=\frac{kq}{r^2}=\frac{9\times10^9\times10^{-6}}{(0.04)^2}=5.625\times10^6\) V/m, directed along +x.
The net field inside a conductor is zero, so the induced field is \(-5.625\times10^6\) V/m.
Answer: (3) \(2.25\times10^{5}\) V and \(-5.625\times10^{6}\) V/m
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