A) \[\frac{b}{a}\]
B) \[\frac{a}{b}\]
C) \[\frac{{{a}^{2}}}{{{b}^{2}}}\]
D) \[\frac{{{b}^{2}}}{{{a}^{2}}}\]
E) \[\frac{(a+b)}{(ab)}\]
Correct Answer: A
Solution :
Given electric potential of spheres are same ie, \[{{V}_{A}}={{V}_{B}}\] \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{Q}_{1}}}{a}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{Q}_{2}}}{b}\] \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{a}{b}\] ?? (i) Surface charge density \[\sigma =\frac{Q}{4\pi {{r}^{2}}}\] \[\Rightarrow \] \[\frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}=\frac{{{Q}_{1}}}{{{Q}_{2}}}\times \frac{{{b}^{2}}}{{{a}^{2}}}\] \[=\frac{a}{b}\times \frac{{{b}^{2}}}{{{a}^{2}}}\] \[=\frac{b}{a}\]You need to login to perform this action.
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