question_answer

Two identical thin rings, each of radius a are placed coaxially at a distance a apart. Let charges \[{{Q}_{1}}\]and \[{{Q}_{2}}\]be placed uniformly on the two rings. The work done in moving a charge q from the centre of one ring to that of the other is

A)  zero       

B) \[\frac{q\sqrt{2}}{4\pi {{\varepsilon }_{0}}a}({{Q}_{1}}-{{Q}_{2}})\]

C) \[\frac{q(\sqrt{2}-1)}{4\pi {{\varepsilon }_{0}}a\sqrt{2}}({{Q}_{1}}-{{Q}_{2}})\]

D) \[\frac{q(\sqrt{2}-1)}{4\pi {{\varepsilon }_{0}}a}({{Q}_{1}}-{{Q}_{2}})\]

Correct Answer: C

Solution :

[c] : The electrostatic potential at the centre of the first ring (i.e., at \[{{Q}_{1}}\]with charge\[{{Q}_{1}}\] is due to charge \[{{Q}_{2}}\]itself as well as due to charge on the second ring which is given by \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}}{a}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{2}}}{a\sqrt{2}}\] Similarly, the electrostatic potential at the centre of the second ring {i.e., at O') is given by \[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}}{a\sqrt{2}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{2}}}{a}\] Required work done ,\[W=q({{V}_{1}}-{{V}_{2}})\] \[=q\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}}{a}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{2}}}{a\sqrt{2}}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}}{a\sqrt{2}}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{2}}}{a} \right]\]\[=\frac{q}{4\pi {{\varepsilon }_{0}}a}\left[ {{Q}_{1}}+\frac{{{Q}_{2}}}{\sqrt{2}}-\frac{{{Q}_{1}}}{\sqrt{2}}-{{Q}_{2}} \right]\] \[=\frac{q}{4\pi {{\varepsilon }_{0}}a}\left[ \frac{\sqrt{2}{{Q}_{1}}+{{Q}_{2}}-{{Q}_{1}}-\sqrt{2}{{Q}_{2}}}{\sqrt{2}} \right]\] \[=\frac{q}{4\pi {{\varepsilon }_{0}}a\sqrt{2}}\left[ \sqrt{2}({{Q}_{1}}-{{Q}_{2}})-1({{Q}_{1}}-{{Q}_{2}}) \right]\] \[=\frac{q(\sqrt{2}-1)}{4\pi {{\varepsilon }_{0}}a\sqrt{2}}({{Q}_{1}}-{{Q}_{2}})\]