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A parallel-plate air condenser of plate area A and separation d is charged to potential V and then the battery is removed. Now a slab of dielectric constant k is introduced between the plates. If Q, E, and W denote respectively the magnitude of the charge on each plate, the electric field between the plates (after the introduction of the dielectric slab) and work done on the system in the process of introducing the slab, then
A \quad W=\frac{\xi_{10} \mathrm{~A} \mathrm{~V} \mathrm{~h}^{2}}{2 \mathrm{~d}}(1-1 / \mathrm{k})
B \quad Q=\frac{\xi_{1} K_{A} V}{d}
c \quad Q=\frac{\xi_{11} \mathrm{~A} \mathrm{~V}}{\mathrm{~d}}
\mathrm{D} \quad \mathrm{E}=\frac{\mathrm{V}}{\mathrm{kd}}

Correct option is A $\mathrm{W}=\frac{\varepsilon_{0} \mathrm{~A} \mathrm{~V} \mathrm{~h}^{2}}{2 \mathrm{~d}}(1-1 / \mathrm{k})$ C $Q=\frac{\varepsilon_{0} A V}{d}$ D $\quad E=\frac{V}{k d}$ As...

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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the space between the plates, is charged to a potential V. The battery is then disconnected and the dielectric slab is withdrawn.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 \mathrm{E}_{1} and \mathrm{E}_{2}, then \mathrm{E}_{1} / \mathrm{E}_{2} is:
A \frac{4}{9}
B \frac{2}{3}
c \frac{3}{2}
D \frac{9}{5}

Correct option is B $\frac{2}{3}$ Let the charge stored by a capacitor with dielectric constant 3 be $Q$. Thus energy stored be $\frac{\mathrm{Q}^{2}}{2 \mathrm{C}_{1}}$ Since the charge remains the...

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