Electrostatics Question 764

Question: Two equally charged spheres of radii a and b are connected together. What will be the ratio of electric field intensity on their surfaces?

Options:

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

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

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

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

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Answer:

Correct Answer: C

Solution:

[c] Let charge on each sphere =q When they are connected together their Now let charge on

$ a=q, $ and on $ b=2q-q _{1} $

$ \Rightarrow V _{a}=V _{b}\text{ or }\frac{1}{4\pi {\varepsilon _{0}}}\frac{q _{1}}{a}=\frac{1}{4\pi {\varepsilon _{0}}}\frac{2q-q _{1}}{b} $

$ \Rightarrow \frac{q _{1}}{2q-q _{1}}=\frac{a}{b} $

$ \frac{E _{a}}{E _{b}}=\frac{\frac{1.}{4\pi {\varepsilon _{0}}}\frac{q _{1}}{a^{2}}}{\frac{1}{4\pi {\varepsilon _{0}}}\frac{q _{2}}{b^{2}}}=( \frac{q _{1}}{2q-q _{1}} )\frac{b^{2}}{a^{2}} $

$ =\frac{a}{b}.\frac{b^{2}}{a^{2}}=\frac{b}{a}=b:a $



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