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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Critical Thinking

  • question_answer
    An ellipsoidal cavity is carved within a perfect conductor. A positive charge \[q\] is placed at the centre of the cavity. The points \[A\] and \[B\] are on the cavity surface as shown in the figure. Then                               [IIT-JEE (Screening) 1999]

    A)            Electric field near \[A\] in the cavity = Electric field near  \[B\] in the cavity

    B)            Charge density at \[A=\] Charge density at \[B\]

    C)            Potential at \[A=\] Potential at \[B\]

    D)            Total electric field flux through the surface of the cavity is \[q/{{\varepsilon }_{0}}\]

    Correct Answer: C , D

    Solution :

     Under electrostatic condition, all points lying on the conductor are in same potential. Therefore, potential at A = potential at B. From Gauss's theorem, total flux through the surface of the cavity will be \[q/{{\varepsilon }_{0}}\]. Not: q Instead of an elliptical cavity, if it would had been a spherical cavity then options (a) and (b) were also correct.

    Solution :

     Under electrostatic condition, all points lying on the conductor are in same potential. Therefore, potential at A = potential at B. From Gauss's theorem, total flux through the surface of the cavity will be \[q/{{\varepsilon }_{0}}\].


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