JEE Main & Advanced Sample Paper JEE Main - Mock Test - 23

  • 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)        \[\frac{2\pi {{\varepsilon }_{0}}R_{1}^{2}R_{2}^{2}{{V}^{2}}}{{{d}^{2}}{{({{R}_{1}}+{{R}_{2}})}^{2}}}\]

    Correct Answer: C

    Solution :

    [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 \[F=K\frac{{{Q}^{2}}}{{{d}^{2}}}=\frac{K}{{{d}^{2}}}{{\left[ \frac{{{R}_{1}}{{R}_{2}}V}{K({{R}_{1}}+{{R}_{2}})} \right]}^{2}}\] \[=\frac{R_{1}^{2}+R_{2}^{2}{{V}^{2}}}{K{{d}^{2}}{{({{R}_{1}}+{{R}_{2}})}^{2}}}=\frac{4\pi {{\varepsilon }_{0}}R_{1}^{2}R_{2}^{2}{{V}^{2}}}{{{d}^{2}}{{({{R}_{1}}+{{R}_{2}})}^{2}}}\]

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