At distance of $5$ $cm$ and $10$ $cm $ outwards from the surface of a uniformly charged solid sphere, the potentials are $100$ $V$ and $75$ $V$ respectively . Then
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Potential at $5 \mathrm{cm}$ from surface $=\frac{K Q}{R+5}=100$
Potential at $10 \mathrm{cm}$ from surface
$=\frac{K Q}{R+10}=75 \Rightarrow R=10 \mathrm{cm}$
$\therefore$ Potential at surface $=\frac{K Q}{R}=\frac{100 \times 15}{10}=150 \mathrm{V}$
Electric field on surface $=\frac{K Q}{R^{2}}=\frac{100 \times 15 V \times c m}{100 c m^{2}}$
$=1500 \mathrm{V} / \mathrm{m}$
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