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(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$,where $A = l 	imes b$ is the area of the plates and $d$ is

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$C$ and $E$ only

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(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$,where $A = l 	imes b$ is the area of the plates and $d$ is the distance between them.Initial capacitance $C_i = \frac{\epsilon_0 (3 \ cm 	imes 1 \ cm)}{3 \ \mu m} = \epsilon_0 	imes 10^4 \ cm^2/m$.We want the new capacitance $C_f = 10 C_i$.For option $A$: $C_A = \frac{\epsilon_0 (30 	imes 1)}{1} = 30 C_i$ (Incorrect).For option $B$: $C_B = \frac{\epsilon_0 (3 	imes 1)}{30} = 0.1 C_i$ (Incorrect).For option $C$: $C_C = \frac{\epsilon_0 (6 	imes 5)}{3} = 10 C_i$ (Correct).For option $D$: $C_D = \frac{\epsilon_0 (1 	imes 1)}{10} = 0.033 C_i$ (Incorrect).For option $E$: $C_E = \frac{\epsilon_0 (5 	imes 2)}{1} = 10 C_i$ (Correct).Thus,options $C$ and $E$ increase the capacitance by a factor of $10$.

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