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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Self Evaluation Test - Electrostatic Potential and Capacitance

  • question_answer
    A capacitor has two circular plates whose radius are 8cm and distance between them is 1mm. When mica (dielectric constant = 6) is placed between the plates. The capacitance of this capacitor and the energy stored when it is given potential of 150 volt respectively are

    A) \[1.06\times {{10}^{-5}}F,1.2\times {{10}^{-9}}J\]

    B) \[1.068\times {{10}^{-9}}F,1.2\times {{10}^{-5}}J\]

    C) \[1.2\times {{10}^{-9}}F,1.068\times {{10}^{-5}}J\]

    D) \[1.6\times {{10}^{-9}}F,1.068\times {{10}^{-5}}J\]

    Correct Answer: B

    Solution :

    [b] Area of plate\[=\pi {{r}^{2}}=\pi \times {{\left( 8\times {{10}^{-2}} \right)}^{2}}\] \[=0.0201{{m}^{2}}\] \[\text{and }d=1mm=1\times {{10}^{-3}}m\] Capacity of capacitor \[C=\frac{{{\varepsilon }_{0}}{{\varepsilon }_{r}}A}{d}=\frac{8.85\times {{10}^{-12}}\times 6\times 0.0201}{1\times {{10}^{-3}}}\] \[=1.068\times {{10}^{-9}}F\] Potential difference, \[V=150\ \text{volt}\] Energy stored, \[U=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\times (1.068\times {{10}^{-9}})\times {{(150)}^{2}}\] \[=1.2\times {{10}^{-5}}J\]


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