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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    A capacitor when filled with a dielectric \[K=3\] has charge \[{{Q}_{0}}\], voltage \[{{V}_{0}}\] and field\[{{E}_{0}}\]. If the dielectric is replaced with another one having \[K=9\] the new values of charge, voltage and field will be respectively

    A)                    \[3{{Q}_{0}},\ 3{{V}_{0}},\ 3{{E}_{0}}\]                 

    B)            \[{{Q}_{0}},\ 3{{V}_{0}},\ 3{{E}_{0}}\]

    C)                    \[{{Q}_{0}},\ \frac{{{V}_{0}}}{3},\ 3{{E}_{0}}\]   

    D)            \[{{Q}_{0}},\ \frac{{{V}_{0}}}{3},\ \frac{{{E}_{0}}}{3}\]

    Correct Answer: D

    Solution :

               When there is no battery, charge remains same while potential difference and electric field decreases i.e. \[Q'={{Q}_{0}},V'=\frac{{{V}_{0}}\times 3}{9}=\frac{{{V}_{0}}}{3}\]and \[E'=\frac{{{E}_{0}}\times 3}{9}=\frac{{{E}_{0}}}{3}\]


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