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 spheres is:

A)  \[\frac{R_{2}^{2}}{R_{2}^{1}}\]

B)  \[\frac{R_{1}^{2}}{R_{2}^{2}}\]

C)  \[\frac{{{R}_{2}}}{{{R}_{1}}}\]

D)  \[\frac{{{R}_{1}}}{{{R}_{2}}}\]

Correct Answer: C

Solution :

Fact: If sphere are joined, they acquire a common potential. If two spheres are joined charge flow till it is uniform both. the spheres, hence electric potential is same. \[\therefore \] \[{{V}_{1}}={{V}_{2}}\] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{{{R}_{1}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{2}}}{{{R}_{2}}}\] \[\Rightarrow \] \[\frac{{{q}_{1}}}{{{R}_{1}}}=\frac{{{q}_{2}}}{{{R}_{2}}}\] ?.(i) Ratio of electric field is given by \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}}{R_{1}^{2}}}{\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{2}}}{R_{2}^{2}}}\] \[\Rightarrow \] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{q}_{1}}}{{{q}_{2}}}{{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{2}}\]                 ?...(ii) Putting the value of \[\frac{{{q}_{1}}}{{{q}_{2}}}\]from equations (i) and (ii), we get \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\,\,{{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{2}}=\frac{{{R}_{2}}}{{{R}_{1}}}\]