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}} $
Show Answer
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} $
BETA
