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

  • question_answer
    Figure given below shows two identical parallel plate capacitors connected to a battery with switch \[S\]closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant 3. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric                                                    [IIT 1983]

    A)                    3 : 1

    B)                    5 : 1

    C)                    3 : 5

    D)                    5 : 3

    Correct Answer: C

    Solution :

               Initially potential difference across both the capacitor is same hence energy of the system is \[{{U}_{1}}=\frac{1}{2}C{{V}^{2}}+\frac{1}{2}C{{V}^{2}}=C{{V}^{2}}\]      ??(i) In the second case when key K is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes 3C, while potential difference across A is V and potential difference across B is \[\frac{V}{3}\] hence energy of the system now is \[{{U}_{2}}=\frac{1}{2}\,(3C){{V}^{2}}+\frac{1}{2}\,(3C)\,{{\left( \frac{V}{3} \right)}^{2}}\]\[=\frac{10}{6}\,C{{V}^{2}}\]                ??(ii) So,  \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{3}{5}\]


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