Electrostatics Question 773

Question: A conducting disc of radius R rotating about its axis with an angular velocity $ \omega $ . Then the potential difference between the center of the disk and its edge is (no magnetic field is present)

Options:

A) zero

B) $ \frac{m _{e}{{\omega }^{2}}R^{2}}{2e} $

C) $ \frac{m _{e}{{\omega }^{2}}R^{3}}{2e} $

D) $ \frac{em _{e}\omega R^{2}}{2} $

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Answer:

Correct Answer: B

Solution:

[b] $ eE=m _{e}{{\omega }^{2}}r $

$ \Rightarrow \int _{{}}^{{}}{Edr=\frac{m _{e}{{\omega }^{2}}r}{e}\int _{0}^{R}{rdr}} $

$ \Rightarrow V=\frac{m _{e}{{\omega }^{2}}R^{2}}{2e} $



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