For the situation shown in the figure below, mark out the correct statement
Medium
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Potential at surface $=$ Potential at centre
$\mathrm{V}_{\mathrm{+q}}+\mathrm{V}_{\mathrm{ind}}=\mathrm{V}_{\mathrm{c}}$
$\mathrm{V}_{\mathrm{ind}}=\frac{\mathrm{Kq}}{(\mathrm{R}+\mathrm{d})}-\frac{\mathrm{Kq}}{\mathrm{d}}=\frac{-\mathrm{qR}}{4 \pi \varepsilon_{0}(\mathrm{d}+\mathrm{R}) \mathrm{d}}$
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