A) \[\frac{2R_{2}^{2}}{R_{1}^{2}}\]
B) \[\frac{R_{1}^{2}}{R_{2}^{2}}\]
C) \[\frac{{{R}_{2}}}{{{R}_{1}}}\]
D) \[\frac{{{R}_{1}}}{{{R}_{2}}}\]
Correct Answer: C
Solution :
It is quite clear, when the two spheres joined by a wire the charge will redistribute and will stop if the potential become equal i.e. \[{{V}_{1}}={{V}_{2}}\] When \[{{q}_{1}}\] and \[{{q}_{2}}\] are charge after redistribution then \[\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{0}}{{R}_{1}}}=\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{0}}{{R}_{2}}}\] \[\frac{{{q}_{1}}}{{{q}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\] Again, \[\frac{{{\varepsilon }_{1}}}{{{\varepsilon }_{2}}}=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{0}}R_{1}^{2}}\times \frac{4\pi {{\varepsilon }_{0}}R_{2}^{2}}{{{q}_{2}}}\] \[\Rightarrow \] \[\frac{{{\varepsilon }_{1}}}{{{\varepsilon }_{2}}}=\frac{{{q}_{1}}}{{{q}_{2}}}\times \frac{R_{2}^{2}}{R_{1}^{2}}\] \[\Rightarrow \] \[\frac{{{\varepsilon }_{1}}}{{{\varepsilon }_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\times \frac{R_{2}^{2}}{R_{1}^{2}}\] \[\Rightarrow \] \[\frac{{{\varepsilon }_{1}}}{{{\varepsilon }_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\]You need to login to perform this action.
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