Electrostatics Question 433

Question: Two spheres A and B of radius a and b respectively are at same electric potential. The ratio of the surface charge densities of A and B is [MP PMT 2001]

Options:

A) $ \frac{a}{b} $

B) $ \frac{b}{a} $

C) $ \frac{a^{2}}{b^{2}} $

D) $ \frac{b^{2}}{a^{2}} $

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Answer:

Correct Answer: B

Solution:

Given electric potential of spheres are same i.e. $ V _{A}=V _{B} $

therefore $ \frac{1}{4\pi {\varepsilon _{0}}}.\frac{Q _{1}}{a}= $ $ \frac{1}{4\pi {\varepsilon _{0}}}.\frac{Q _{2}}{b}\Rightarrow $ $ \frac{Q _{1}}{Q _{2}}=\frac{a}{b} $ ……(i)

as surface charge density $ \sigma =\frac{Q}{4\pi r^{2}} $

therefore $ \frac{{\sigma _{1}}}{{\sigma _{2}}}=\frac{Q _{1}}{Q _{2}}\times \frac{b^{2}}{a^{2}}=\frac{a}{b}\times \frac{b^{2}}{a^{2}}=\frac{b}{a} $



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