A parallel plate air capacitor of capacitance | vedclass

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(D) Initially,the capacitor is charged to $q = CV$. When the cell is disconnected,the charge $q$ remains constant on the plates.After inserting the di

Vedclass · Accepted Answer

The charge on the capacitor is not conserved.

Vedclass · Answer

(D) Initially,the capacitor is charged to $q = CV$. When the cell is disconnected,the charge $q$ remains constant on the plates.After inserting the dielectric slab,the new capacitance becomes $C' = KC$.The potential difference between the plates becomes $V' = \frac{q}{C'} = \frac{CV}{KC} = \frac{V}{K}$. Thus,the potential difference decreases $K$ times.The initial energy stored is $U_1 = \frac{q^2}{2C} = \frac{1}{2}CV^2$.The final energy stored is $U_2 = \frac{q^2}{2C'} = \frac{q^2}{2KC} = \frac{U_1}{K}$. Thus,the energy decreases $K$ times.The change in energy is $\Delta U = U_2 - U_1 = \frac{q^2}{2KC} - \frac{q^2}{2C} = \frac{1}{2}CV^2\left(\frac{1}{K} - 1\right)$.Since the capacitor is disconnected from the cell,the charge $q$ on the plates remains constant (conserved). Therefore,the statement that the charge is not conserved is incorrect.

$A$ parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. $A$ dielectric slab of dielectric constant $K,$ which can just fill the air gap of the capacitor,is now inserted in it. Which of the following is incorrect $?$

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