• # 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)  $\frac{{{R}_{1}}}{{{R}_{2}}}$

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