A parallel plate capacitor with vacuum between the plates has a capacitance of 3
ID: 1491183 • Letter: A
Question
A parallel plate capacitor with vacuum between the plates has a capacitance of 3.547 mF. A dielectric material with k=4.617 is placed between the plates completely filling the volume between them . The capacitor is then connected to a battery that maintains a potential difference of 10.03 V across the plates. How much work is required to pull the dielectric material out of the capacitor? A parallel plate capacitor with vacuum between the plates has a capacitance of 3.547 mF. A dielectric material with k=4.617 is placed between the plates completely filling the volume between them . The capacitor is then connected to a battery that maintains a potential difference of 10.03 V across the plates. How much work is required to pull the dielectric material out of the capacitor?Explanation / Answer
required work is W = U1-U2
U1 is the energy stored with dilectric = 0.5*C'*V^2 = 0.5*(4.617*3.547*10^-3)*10.03^2 = 0.823 J
U2 is the energy stored without dielectric = 0.5*C*V^2 = 0.4*3.547*10^-3*10.03^2 =0.142 J
then W = 0.823-0.142= 0.681 J