A) \[a/b\]
B) \[b/a\]
C) \[{{a}^{2}}/{{b}^{2}}\]
D) \[{{b}^{2}}/a\]
Correct Answer: B
Solution :
Joined by a wire means they are at the same potential. For same potential \[\frac{k{{Q}_{1}}}{{{a}_{1}}}=\frac{k{{Q}_{2}}}{{{a}_{2}}}\]Þ \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{a}{b}\] Further, the electric field at the surface of the sphere having radius R and charge Q is \ \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{k{{Q}_{1}}/{{a}^{2}}}{k{{Q}_{2}}/{{b}_{2}}}=\frac{{{Q}_{1}}}{{{Q}_{2}}}\times \frac{{{b}^{2}}}{{{a}^{2}}}=\frac{b}{a}\]You need to login to perform this action.
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