question_answer

A thin metallic disc of radius R is rotating with constant angular velocity co about a vertical axis that is perpendicular to plane of disc and passes through its centre.

Rotation causes free electrons in the disc to redistribute. If there is no external electric or magnetic field is present, then potential difference between centre and rim of disc is

A) \[0\]

B) \[\frac{m{{\omega }^{2}}{{R}^{2}}}{e}\]

C) \[\frac{m{{\omega }^{2}}{{R}^{2}}}{2e}\]

D) \[\frac{2m{{\omega }^{2}}{{R}^{2}}}{e}\]

Correct Answer: C

Solution :

In rotating disc, let a potential difference dV occurs across radial distance dr.

In equilibrium, electric field balances centrifugal   force on an electron.

\[\therefore \]\[e\frac{dV}{dr}=m{{\omega }^{2}}r\]\[\Rightarrow \]\[dV=\frac{m}{e}{{\omega }^{2}}rdr\]

\[\therefore \]Integrating between limits, we have, \[V=\int_{0}^{V(R)}{dV=\int_{0}^{R}{\frac{m}{e}{{\omega }^{2}}rdr}}\]

\[V(R)=\frac{-m{{\omega }^{2}}{{R}^{2}}}{e\cdot 2}\]

Note That outer rim will be negatively charged In practice the potential difference is very small.