A parallel plate capacitor with a dielectric slab of dielectric constant $3$ , filling the space between the plates, is charged to a potential $V$ . The battery is then disconnected and the dielectric slab is withdrawn. It is then replaced by another dielectric slab of dielectric constant $2$ . If the energies stored in the capacitor before and after the dielectric slab is changed are $E_{1}$ and $E_{2}$ , then $E_{1} / E_{2}$ is:

\frac{4}{9}

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\frac{2}{3}

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\frac{3}{2}

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\frac{9}{5}

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Solution

The correct option is B $\frac{2}{3}$
Let the charge stored by capacitor with dielectric constant $3$ be $Q$ .
Thus energy stored be $\frac{Q^{2}}{2 C_{1}}$

Since the charge remains the same after changing the dielectric, the new energy stored= $\frac{Q^{2}}{2 C_{2}}$

$⟹ \frac{E_{1}}{E_{2}} = \frac{C_{2}}{C_{1}} = \frac{K_{2}}{K_{1}} = \frac{2}{3}$

$(C = \frac{K A ϵ_{0}}{d})$

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The correct option is B 23Let the charge stored by capacitor with dielectric constant 3 be Q.Thus energy stored be Q22C1Since the charge remains the same after changing the dielectric, the new energy stored=Q22C2⟹E1E2=C2C1=K2K1=23 (C=KAϵ0d)