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} ]. $



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