question_answer

Two spheres of radii \[{{R}_{1}}\] and \[{{R}_{2}}\] respectively are charged and joined by a wire. The ratio of electric field of sphere will be:

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,       \[\frac{{{\varepsilon }_{1}}}{{{\varepsilon }_{2}}}=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{0}}R_{1}^{2}}\times \frac{4\pi {{\varepsilon }_{0}}R_{2}^{2}}{{{q}_{2}}}\] \[\Rightarrow \]            \[\frac{{{\varepsilon }_{1}}}{{{\varepsilon }_{2}}}=\frac{{{q}_{1}}}{{{q}_{2}}}\times \frac{R_{2}^{2}}{R_{1}^{2}}\] \[\Rightarrow \]            \[\frac{{{\varepsilon }_{1}}}{{{\varepsilon }_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\times \frac{R_{2}^{2}}{R_{1}^{2}}\] \[\Rightarrow \]            \[\frac{{{\varepsilon }_{1}}}{{{\varepsilon }_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\]