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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Mock Test - Capacitance

  • question_answer
    A parallel plate capacitor of plate area A and plate separation d is charged to potential V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitors so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and work done on the system in question in the process of inserting the slab, then state incorrect relation from the following

    A) \[Q=\frac{{{\varepsilon }_{0}}AV}{d}\]

    B) \[W=\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{2Kd}\]

    C) \[E=\frac{{{V}^{{}}}}{Kd}\]  

    D) \[W=\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{2d}\left( 1-\frac{1}{K} \right)\]  

    Correct Answer: D

    Solution :

    [d] After inserting the dielectric slab New capacitance \[C'=K.C=\frac{K{{\varepsilon }_{0}}A}{d}\] New potential difference \[V'=\frac{V}{K}\] New charge \[Q'=C'V'=\frac{{{\varepsilon }_{0}}AV}{d}\] Mew electric field \[E'=\frac{V'}{d}=\frac{V}{Kd}\] Work done (W) = final energy - initial energy \[W=\frac{1}{2}C'V{{'}^{2}}-\frac{1}{2}C{{V}^{2}}=\frac{1}{2}(KC){{\left( \frac{V}{K} \right)}^{2}}-\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}C{{V}^{2}}\left( \frac{1}{K}-1 \right)=-\frac{1}{2}C{{V}^{2}}=\left( 1-\frac{1}{K} \right)\] \[=\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{2d}\left( 1-\frac{1}{K} \right)\]so \[\left| W \right|=\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{2d}\left( 1-\frac{1}{K} \right)\]


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