The electric potential at the surface of an atomic nucleus $(z=50)$ of radius $9 \times 10^{-13} \mathrm{~cm}$ is __________ $\times 10^6 \mathrm{~V}$.
Answer (integer)
8
Solution
<p>$$\begin{aligned}
& \text { Potential }=\frac{\mathrm{kQ}}{\mathrm{R}}=\frac{\mathrm{k} . \mathrm{Ze}}{\mathrm{R}} \\
& =\frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13} \times 10^{-2}} \\
& =8 \times 10^6 \mathrm{~V}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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