What are the new values of V, Q, U, and E?

An air-filled parallel plate capacitor is connected to a battery of voltage #V_0#. There is charge #Q_0# on each plate,
energy #U_0# stored between the plates and an electric field of #E_0# between the plates. The capacitor remains connected to the battery and a dielectric material with a dielectric constant of #kappa# is then placed between the plates of the capacitor. What are the new values of V, Q, U, and E?

1 Answer
Apr 18, 2018

#V = V_0,quad Q = kappa Q_0,quad U = kappa U_0,quad E = E_0 #

Explanation:

The capacitance rises by a factor of #kappa# as a result of the introduction of the dielectric: #C=kappa C_0#.

Since the capacitor remains connected to the battery #V=V_0#.

#Q = CV = kappa C_0 V_0 = kappa Q_0#

#U = 1/2 CV^2 = 1/2 kappa C_0 V_0^2 = kappa U_0#

#E = V/d = V_0/d = E_0#