2. Electric Potential and Capacitance
hard

The electric potential at the surface of an atomic nucleus $(z=50)$ of radius $9 \times 10^{-13} \mathrm{~cm}$ is ________$\times 10^6 \mathrm{~V}$.

A

$10$

B

$9$

C

$7$

D

$8$

(JEE MAIN-2024)

Solution

$ \text { Potential }=\frac{k Q}{R}=\frac{k \cdot Z e}{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}$

Standard 12
Physics

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