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 \[b=2q-{{q}_{1}}\] \[\Rightarrow {{V}_{a}}={{V}_{b}}\text{ or }\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{a}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2q-{{q}_{1}}}{b}\] \[\Rightarrow \frac{{{q}_{1}}}{2q-{{q}_{1}}}=\frac{a}{b}\] \[\frac{{{E}_{a}}}{{{E}_{b}}}=\frac{\frac{1.}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{{{a}^{2}}}}{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{2}}}{{{b}^{2}}}}=\left( \frac{{{q}_{1}}}{2q-{{q}_{1}}} \right)\frac{{{b}^{2}}}{{{a}^{2}}}\] \[=\frac{a}{b}.\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{b}{a}=b:a\]You need to login to perform this action.
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