Electrostatics Question 771
Question: A plastic disc is charged on one side with a uniform surface charge density $ \sigma $ and then three quadrant of the disk are removed. The remaining quadrant , with V=0 at infinity, the potential due to the remaining quadrant point P is
Options:
A) $ \frac{\sigma }{2{\in _{0}}}[ {{( r^{2}+R^{2} )}^{1/2}}-r ] $
B) $ \frac{\sigma }{2{\in _{0}}}[ R-r ] $
C) $ \frac{\sigma }{8{\in _{0}}}[ {{( r^{2}+R^{2} )}^{1/2}}-r ] $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] The potential at P due to whole disc is
$ V=\frac{\sigma }{2{\in _{0}}}[ \sqrt{R^{2}+r^{2}-r} ]. $
Now potential due to quarter disc,
$ V=\frac{V}{4}=\frac{\sigma }{8{\in _{0}}}[ \sqrt{R^{2}+r^{2}-r} ]. $
