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} $
Show Answer
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} $
