NEET Sample Paper NEET Sample Test Paper-15

  • question_answer There are two hollow concentric spheres with radii \[{{R}_{1}}\]and R for the inner and outer spheres respectively. The inner sphere is given a charge \[{{Q}_{1}}\]and the outer a charge\[{{Q}_{2}}.\]Then the potential at the surface of the inner sphere is

    A)  \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}}+\frac{{{Q}_{2}}}{{{R}_{2}}} \right)\]

    B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}}-\frac{{{Q}_{2}}}{{{R}_{2}}} \right)\]

    C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}} \right)\]                      

    D)  0

    Correct Answer: A

    Solution :

    Potential at the surface of the inner sphere due to the charges of the outer sphere \[={{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{2}}}{{{R}_{2}}}\] Potential at the surface of the inner sphere due to its own charges \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}}{{{R}_{1}}}\] .'. Net potential on the surface of the inner sphere \[=V={{V}_{1}}+{{V}_{2}}\] or,        \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}}+\frac{{{Q}_{2}}}{{{R}_{2}}} \right)\] Hence, the correction option is [a].


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