A parallel plate capacitor is given a charge Q. After charging, the capacitor is
ID: 1589581 • Letter: A
Question
A parallel plate capacitor is given a charge Q. After charging, the capacitor is removed from the circuit so that the charge remains fixed. The potential energy stored on the capacitor is
PE = 1/2QV = 1/2Q2/C = 1/2CV2.
A student inserts a dielectric into the space between the plates, effectively reducing the electric field between the plates by a factor equal to the dielectric constant K. This increases the capacitance by a factor K.
1. Has the electrical energy stored in the capacitor increased or decreased? Defend your answer.
Explanation / Answer
electrical energy stored is U = 0.5*C*V^2
if capacitance is increased by a factor k
then energy is U'= 0.5*(k*C)*V^2
U'= k*0.5*C*V^2 = k*U
i.e the energy stored also increases by factor k