JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • question_answer
    A charge Q is distributed over three concentric spherical shells of radii a, b, c \[\left( a<b<c \right)\] such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where \[r<a\], would be- [JEE Main 10-Jan-2019 Morning]

    A) \[\frac{Q({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}{{{4}_{\pi {{\varepsilon }_{0}}}}({{a}^{3}}+{{b}^{3}}+{{c}^{3}})}\]

    B) \[\frac{Q}{{{4}_{\pi {{\varepsilon }_{0}}}}(a+b+c)}\]

    C) \[\frac{Q}{{{12}_{\pi \varepsilon {{\,}_{0}}}}}\,\,\frac{ab+bc+ca}{abc}\]

    D) \[\frac{Q(a+b+c)}{{{4}_{\pi \varepsilon {{\,}_{0}}}}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}\]

    Correct Answer: D

    Solution :

    [d]
     
    \[x+y+z=Q\]
    \[\frac{x}{4\pi {{a}^{2}}}=\frac{y}{4\pi {{b}^{2}}}=\frac{z}{4\pi {{c}^{2}}}\]
    (surface charge density for all are same)
    \[x:y:z={{a}^{2}}:{{b}^{2}}:{{c}^{2}}\]
    \[x=\frac{{{a}^{2}}Q}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}};\,\,\,y\,=\,\frac{{{b}^{2}}Q}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}};\]
    \[z=\frac{{{c}^{2}}Q}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]
    V at \[r<a\] \[=\,\frac{kx}{a}+\frac{ky}{b}+\frac{kz}{c}\]
    \[=\frac{1}{4\pi {{\in }_{0}}}\left[ \frac{Qa}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\,+\,\frac{Qb}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\,+\,\,\frac{Qc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \right]\]\[=\,\,\frac{Q}{4\pi {{\in }_{0}}}\,\,\left[ \frac{a+b+c}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \right]\]


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