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A parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to potential difference $V$ and then the battery is disconnected. A slab of dielectric constant $K$ is then inserted between the plates of the capacitor 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 questions, in the process of inserting the slab, then

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$Q = \frac{ε_{0} A V}{d}$

$Q = \frac{ε_{0} K A V}{d}$

$E = \frac{V}{K d}$

$W = - \frac{ε_{0} A V^{2}}{2 d} (1 - \frac{1}{K})$

answer is A, C, D.

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Detailed Solution

Let $C_{0}$ be the capacitance initially and $C$ be the capacitance finally. Then $C_{0} = \frac{ε_{0} A}{d}$

Since, $Q = C_{0} V$

$\Rightarrow Q = \frac{ε_{0} A V}{d}$

Further $E_{0} = \frac{V}{d}$ and $E = \frac{E_{0}}{K}$

$\Rightarrow E = \frac{V}{K d}$

Also, if $U_{i}$ is the initial energy, then $U_{i} = \frac{1}{2} C_{0} V^{2}$

After the introduction of slab if $U_{f}$ be the final energy then

$U_{f} = \frac{1}{2} C V_{slab}^{2} = \frac{1}{2} (K C_{0}) {(\frac{V}{K})}^{2} \Rightarrow U_{f} = \frac{1}{2} \frac{C_{0} V^{2}}{K} \Rightarrow Δ U = U_{2} - U_{1} \Rightarrow Δ U = \frac{1}{2} C_{0} V^{2} (\frac{1}{K} - 1)$

Since work done = Decrease in Potential Energy

$\Rightarrow W = Δ U \Rightarrow W = - \frac{1}{2} \frac{ε_{0} A V^{2}}{d} (1 - \frac{1}{K})$

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