A parallel-plate air capacitor of area A = 19.9 cm 2 and plate separation of d =

Question

A parallel-plate air capacitor of area A = 19.9 cm2 and plate separation of d = 3.10 mm is charged by a battery to a voltage of 56.0 V. What is the charge on the capacitor?

If a dielectric material with ? = 4.70 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate?

Explanation / Answer

Q = C*V => Q = e0*A*V/d => Q= 8.854*10^-12*19.9*10^-4*56/3.1*10^-3 = 3.183*10^-10 C when dielectric is inserted,the potential of a capacitor decreases by a factor of K when a dielectric material is added and the charge on the capacitor stays constant. But since the capacitor is still connected to the battery, V = 56 V additional charge flown = C' *(V - V/4.7) = 4.7*C*(56 - 56/4.7) = [4.7*8.854*10^-12*19.9*10^-4/3.1*10^-3]*44.09 = 1.178*10^-9 C = 1.178 nC

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