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

  • question_answer
    In the rectangle, shown below, the two corners have charges \[{{q}_{1}}=-5\mu C\] and \[{{q}_{2}}=+2.0\mu C\]. The work done in moving a charge \[+3.0\mu C\] from \[B\] to \[A\] is (take \[1/4\pi {{\varepsilon }_{0}}={{10}^{10}}N\text{-}{{m}^{2}}/{{C}^{2}}\])                                             [AMU 2001]

    A)            2.8 J

    B)                                      3.5 J

    C)            4.5 J                                         

    D)            5.5 J

    Correct Answer: A

    Solution :

                         Work done \[W=3\times {{10}^{-6}}({{V}_{A}}-{{V}_{B}});\] where \[{{V}_{A}}={{10}^{10}}\left[ \frac{(-5\times {{10}^{-6}})}{15\times {{10}^{-2}}}+\frac{2\times {{10}^{-6}}}{5\times {{10}^{-2}}} \right]=\frac{1}{15}\times {{10}^{6}}\,volt\] and \[{{V}_{B}}={{10}^{10}}\left[ \frac{(2\times {{10}^{-6}})}{15\times {{10}^{-2}}}-\frac{5\times {{10}^{-6}}}{5\times {{10}^{-2}}} \right]=-\frac{13}{15}\times {{10}^{6}}\,volt\] \ \[W=3\times {{10}^{-6}}\left[ \frac{1}{15}\times {{10}^{6}}-\left( -\frac{13}{15}\times {{10}^{6}} \right) \right]\]= 2.8 J


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