JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    Two insulated charged conducting spheres of radii \[20\,cm\] and \[15\,cm\]respectively and having an equal charge of \[10\,C\] are connected by a copper wire and then they are separated. Then                                                [MP PET 1997]

    A)                    Both the spheres will have the same charge of \[10\,C\]

    B)                    Surface charge density on the \[20\,cm\] sphere will be greater than that on the \[15\,cm\] sphere

    C)                    Surface charge density on the \[15\,cm\] sphere will be greater than that on the \[20\,cm\] sphere

    D)                    Surface charge density on the two spheres will be equal

    Correct Answer: C

    Solution :

                         After redistribution, charges on them will be different, but they will acquire common potential i.e. \[k\frac{{{Q}_{1}}}{{{r}_{1}}}=k\frac{{{Q}_{2}}}{{{r}_{2}}}\] Þ \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\] As \[\sigma =\frac{Q}{4\pi \,{{r}^{2}}}\] Þ \[\frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}=\frac{{{Q}_{1}}}{{{Q}_{2}}}\times \frac{r_{2}^{2}}{r_{1}^{2}}\] Þ \[\frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}=\frac{{{r}_{2}}}{{{r}_{1}}}\] Þ \[\sigma \propto \frac{1}{r}\] i.e. surface charge density on smaller sphere will be more.


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