Discussion
The potential at the surface of a conducting sphere of radius 0.1 m carrying 5 nC is
Answer: B
Correct answer: 450 V
\[ V = \frac{Q}{4\pi\varepsilon_0 r} = \frac{5\times10^{-9}}{4\pi \times 8.854\times10^{-12} \times 0.1} = 449\ \mathrm{V}. \] Outside a conducting sphere the field and potential are those of a point charge at its centre, which is why the same formula serves.
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