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Q. The electric potential at the surface of an atomic nucleus (Z = 50) of radius $ 9 \times 10^{ - 15} \, m$ is

Gujarat CETGujarat CET 2006Electrostatic Potential and Capacitance

Solution:

Electric potential at surface,
V = $ \frac{ 1}{ 4 \pi \varepsilon_0 } \frac{ q}{ R} ( q = Ze)$
= $ 9 \times 10^9 \times \frac{ 50 \times 1.6 \times 10^{ - 19}}{ 9 \times 10^{ - 15}} $
= $ 8 \times 10^6 $ Volt