A) \[\frac{a}{b}\]
B) \[\frac{b}{a}\]
C) \[\frac{{{a}^{2}}}{{{b}^{2}}}\]
D) \[\frac{{{b}^{2}}}{{{a}^{2}}}\]
Correct Answer: B
Solution :
Given, electric potentials of spheres are same ie, \[{{V}_{A}}={{V}_{B}}\] \[\Rightarrow \] \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}}{a}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{2}}}{b}\] \[\Rightarrow \] \[\frac{{{q}_{1}}}{{{q}_{2}}}=\frac{a}{b}\] ?.(i) As surface charge density \[\sigma =\frac{q}{4\pi \,{{r}^{2}}}\] \[\Rightarrow \] \[\frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}=\frac{{{q}_{1}}}{{{q}_{2}}}\times \frac{{{b}^{2}}}{{{a}^{2}}}\] \[=\frac{a}{b}\times \frac{{{b}^{2}}}{{{a}^{2}}}=\frac{b}{a}\]
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