• # question_answer Two solid conducting spheres of radii ${{R}_{1}}$ and ${{R}_{2}}$are kept at a distance d ($>>{{R}_{1}}$ and ${{R}_{2}}$) apart. The two spheres are connected by thin conducting wires to the positive and negative terminals of a battery of emf V. Find the electrostatic force between the two spheres. A) $\frac{\pi {{\varepsilon }_{0}}R_{1}^{2}R_{2}^{2}{{V}^{2}}}{{{d}^{2}}{{({{R}_{1}}+{{R}_{2}})}^{2}}}$ B)        $\frac{\pi {{\varepsilon }_{0}}R_{1}^{2}R_{2}^{2}{{V}^{2}}}{4{{d}^{2}}{{({{R}_{1}}+{{R}_{2}})}^{2}}}$ C) $\frac{4\pi {{\varepsilon }_{0}}R_{1}^{2}R_{2}^{2}{{V}^{2}}}{{{d}^{2}}{{({{R}_{1}}+{{R}_{2}})}^{2}}}$ D)        $\frac{2\pi {{\varepsilon }_{0}}R_{1}^{2}R_{2}^{2}{{V}^{2}}}{{{d}^{2}}{{({{R}_{1}}+{{R}_{2}})}^{2}}}$

[c] Let's first calculate the capacitance of the system. Assuming that the two spheres have charges Q and $-Q,$we can write their potentials as ${{V}_{1}}=\frac{KQ}{{{R}_{1}}}$ and  ${{V}_{2}}=\frac{K(-Q)}{{{R}_{2}}}$ Note that charge on one sphere does not have any significant effect on the potential of other since $d>>{{R}_{1}}$and ${{R}_{2}}$. $\therefore$  Potential difference between the two spheres $V={{V}_{1}}-{{V}_{2}}=\frac{KQ}{{{R}_{1}}}+\frac{KQ}{{{R}_{2}}}$ $\therefore$   Capacitance $C=\frac{Q}{V}=\frac{{{R}_{1}}{{R}_{2}}}{K({{R}_{1}}+{{R}_{2}})}$ Charge on the spheres $Q=\frac{{{R}_{1}}{{R}_{2}}V}{\left( {{R}_{1}}+{{R}_{2}} \right)}$ $\therefore$  Force between the spheres $F=K\frac{{{Q}^{2}}}{{{d}^{2}}}=\frac{K}{{{d}^{2}}}{{\left[ \frac{{{R}_{1}}{{R}_{2}}V}{K({{R}_{1}}+{{R}_{2}})} \right]}^{2}}$ $=\frac{R_{1}^{2}+R_{2}^{2}{{V}^{2}}}{K{{d}^{2}}{{({{R}_{1}}+{{R}_{2}})}^{2}}}=\frac{4\pi {{\varepsilon }_{0}}R_{1}^{2}R_{2}^{2}{{V}^{2}}}{{{d}^{2}}{{({{R}_{1}}+{{R}_{2}})}^{2}}}$