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Q. Two spheres of radius $ {{R}_{1}} $ and $ {{R}_{2}} $ respectively are charged and joined by a wire. The ratio of electric field of the spheres is:

JIPMERJIPMER 2000

Solution:

When two spheres are joined by the wire by doing so we observe that, they are at same potential and for same potential $ \frac{K{{q}_{1}}}{{{R}_{1}}}=\frac{K{{q}_{2}}}{{{R}_{2}}} $ or $ \frac{{{q}_{1}}}{{{q}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}} $ Again the electric field at the surface of the sphere having radius R and charge q is given by $ E=\frac{Kq}{{{R}^{2}}} $ Hence $ \frac{{{E}_{1}}}{{{E}_{2}}}=\frac{\frac{K{{q}_{1}}}{R_{1}^{2}}}{\left( \frac{Kq_{2}^{2}}{R_{2}^{2}} \right)}=\frac{{{q}_{1}}}{{{q}_{2}}}\times \frac{R_{2}^{2}}{R_{1}^{2}} $ $ =\frac{{{R}_{1}}}{{{R}_{2}}}\times \frac{R_{2}^{2}}{R_{1}^{2}}=\frac{{{R}_{2}}}{{{R}_{1}}}. $