JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Critical Thinking

  • question_answer
    nnCapacitance of a capacitor made by a thin metal foil is \[2\mu F\]. If the foil is folded with paper of thickness\[0.15mm\], dielectric constant of paper is 2.5 and width of paper is\[400mm\], then length of foil will be                   [RPET 1997]

    A)                    \[0.34\ m\]

    B)                                      \[1.33\ m\]

    C)                    \[13.4\ m\]                    

    D)            \[33.9\ m\]

    Correct Answer: D

    Solution :

                       If length of the foil is l then \[C=\frac{k{{\varepsilon }_{0}}(l\times b)}{d}\] Þ \[2\times {{10}^{-6}}=\frac{2.5\times 8.85\times {{10}^{-12}}(l\times 400\times {{10}^{-3}})}{0.15\times {{10}^{-3}}}\] Þ l = 33.9 m


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