A parallel plate capacitor with air between the plates has a capacitance of $8 ext{ pF}$. What will

Question

A parallel plate capacitor with air between the plates has a capacitance of $8	ext{ pF}$. What will be the capacitance if the distance between the plates is reduced by half, and the space is filled with a substance of dielectric constant 6?

Professor PC Thomas Classes · Answer

Initial capacitance $C_0 = \frac{\epsilon_0 A}{d} = 8	ext{ pF}$. New capacitance $C = \frac{K\epsilon_0 A}{d/2} = 2K \left(\frac{\epsilon_0 A}{d}ight) = 2 	imes 6 	imes 8	ext{ pF} = 96	ext{ pF}$.