A parallel plate capacitor of capacitance 5,m | vedclass

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(C) Initial charge on the capacitor plates without the dielectric is:$Q = CV = (5\,\mu F) 	imes (1\,V) = 5\,\mu C$The capacitance $C'$ after introduc

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(C) Initial charge on the capacitor plates without the dielectric is:$Q = CV = (5\,\mu F) 	imes (1\,V) = 5\,\mu C$The capacitance $C'$ after introducing a dielectric slab of thickness $t$ and dielectric constant $K$ in a capacitor with plate separation $d$ is given by:$C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} = \frac{C}{1 - \frac{t}{d}(1 - \frac{1}{K})}$Substituting the given values $(C = 5\,\mu F, d = 6\,cm, t = 4\,cm, K = 4)$:$C' = \frac{5\,\mu F}{1 - \frac{4}{6}(1 - \frac{1}{4})} = \frac{5\,\mu F}{1 - \frac{2}{3}(\frac{3}{4})} = \frac{5\,\mu F}{1 - \frac{1}{2}} = \frac{5\,\mu F}{0.5} = 10\,\mu F$The new charge on the capacitor plates is:$Q' = C'V = (10\,\mu F) 	imes (1\,V) = 10\,\mu C$The additional charge that flows from the battery is:$\Delta Q = Q' - Q = 10\,\mu C - 5\,\mu C = 5\,\mu C$

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