A parallel plate capacitor with vacuum between the plates has a capacitance of 3

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

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