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

  • question_answer
    There are two equipotential surface as shown in figure. The distance between them is r. The charge of ?q coulomb is taken from the surface A to B, the resultant work done will be                   [MP PMT 1986; CPMT 1986, 88]

    A)                    \[W=\frac{1}{4\pi {{\varepsilon }_{o}}}\frac{q}{r}\]

    B)                    \[W=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}\]

    C)                    \[W=-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}\]

    D)                    W = zero

    Correct Answer: D

    Solution :

               The work done  is given by \[=q({{V}_{2}}-{{V}_{1}})=0\]


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