A parallel plate capacitor of capacity 5µF and plate separation 6cm is connected to a 1V battery and is charged. A dielectric of dielectric constant 4 and thickness 4 cm introduced into the capacitor. The additional charge that flows into the capacitor from the battery is

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3μC

10μC

2μC

5μC

answer is C.

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Detailed Solution

$\begin{array}{l} C_{0} = \frac{{Aε}_{0}}{d}; 5 \times 10^{- 6} = \frac{{Aε}_{0}}{6 \times 10^{- 2}} \\ \Rightarrow {Aε}_{0} = 30 \times 10^{- 8} \\ Q_{1} = C_{0} V = 5 \times 10^{- 6} \times 1 = 5 μC \\ C = \frac{{Aε}_{0}}{d - t (1 - \frac{1}{K})} = \frac{{Aε}_{0}}{6 - 4 (1 - \frac{1}{4})} \\ C = \frac{{Aε}_{0}}{3 \times 10^{- 2}} = \frac{30 \times 10^{- 8}}{3 \times 10^{- 2}} = 10 μF \\ Q_{2} = CV = 10 \times 10^{- 6} \times 1 = 10 μC \end{array}$
Additional charge =Q₂–Q₁ = 10-5 = 5µC