question_answer

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

A) \[\frac{a}{b}\]

B) \[\frac{{{a}^{2}}}{{{b}^{2}}}\]

C) \[\frac{b}{a}\]

D) \[\frac{{{b}^{2}}}{{{a}^{2}}}\]

Correct Answer: C

Solution :

[c] Let charge on each sphere =q When they are connected together their Now let charge on \[a\text{ }=q,\] and on \[b=2q-{{q}_{1}}\] \[\Rightarrow {{V}_{a}}={{V}_{b}}\text{ or }\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{a}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2q-{{q}_{1}}}{b}\] \[\Rightarrow \frac{{{q}_{1}}}{2q-{{q}_{1}}}=\frac{a}{b}\] \[\frac{{{E}_{a}}}{{{E}_{b}}}=\frac{\frac{1.}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{{{a}^{2}}}}{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{2}}}{{{b}^{2}}}}=\left( \frac{{{q}_{1}}}{2q-{{q}_{1}}} \right)\frac{{{b}^{2}}}{{{a}^{2}}}\] \[=\frac{a}{b}.\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{b}{a}=b:a\]