The electric potential at the surface of an atomic nucleus (... | Physics | SolveeIt

Question

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

80 V

$8 \times 10^6$ V

9 V

$9 \times 10^5$ V

Answer

$8 \times 10^6$ V

Explanation

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

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Solution