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

  • question_answer
    If a charge of \[+2.0\times {{10}^{-8}}C\]is placed on the positive plate and a charge of \[-1.0\times {{10}^{-8}}C\]on the negative plate of a parallel plate capacitor of capacitance\[1.2\times {{10}^{-3}}\mu F\], then the potential difference developed between the plates is:

    A) 6.25V                           

    B) 3.0V     

    C) 12.5 V                          

    D) 25V

    Correct Answer: C

    Solution :

    (c) 12.5V  Here, \[V=\frac{{{q}_{1}}-{{q}_{2}}}{2C}\] \[=\frac{2.0\times {{10}^{-8}}+1.0\times {{10}^{-8}}}{2\times 1.2\times {{10}^{-9}}}=12.5V\]


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