12th Class Physics Electrostatics & Capacitance Question Bank Case Based (MCQs) - Electric Potential and Capacitance

  • question_answer
    The point charges A = +3 nC and B = +1 nC are placed 5 cm apart in air. The work done to move charge B towards A by 1 cm is

    A) \[2\centerdot 0\times {{10}^{-7}}\,J\]

    B) \[1\centerdot 35\times {{10}^{-7}}\,J\]

    C) \[2\centerdot 7\times {{10}^{-7}}\,J\]

    D) \[1\centerdot 21\times {{10}^{-7}}\,J\]

    Correct Answer: B

    Solution :

    Required work done = Change in potential energy of the system \[W={{U}_{f}}-{{U}_{i}}=k\frac{{{q}_{1}}{{q}_{2}}}{{{r}_{f}}}-k\frac{{{q}_{1}}{{q}_{2}}}{{{r}_{i}}}\] \[=k\,{{q}_{1}}{{q}_{2}}\left[ \frac{1}{{{r}_{f}}}-\frac{1}{{{r}_{i}}} \right]\] \[\therefore \,\,W=\left( 9\times {{10}^{9}} \right)\left( 3\times {{10}^{-9}}\times 1\times {{10}^{-9}} \right)\] \[\left[ \frac{1}{4\times {{10}^{-2}}}-\frac{1}{5\times {{10}^{-2}}} \right]\] \[=27\times {{10}^{-7}}\times \left( 0\,.\,05 \right)=1\,.\,35\times {{10}^{-7}}J\]


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