A 2nF parallel-plate capacitor with mica (kappa = 5) in between the plates is ch

Question

A 2nF parallel-plate capacitor with mica (kappa = 5) in between the plates is charged to a potential difference of 100 V and the disconnected from the source. How much work is required to pull mica out? After mica has been pulled out, Find C, deltaV, and, Q. Find change in potential energy of the capacitor - with and without mica. Explain the change in energy. Suppose the capacitor is kept connected to the power supply while is removed from between the plates. Answer (b) and (c) for this scenario.

Explanation / Answer

a) C with mica = 2 nF

Q = CV = 2*100 = 200 nC

kappa of mica = 5

V = 100 V

energy with mica = Q2/2C = 2002 / 2*2 = 10000 nJ

C without mica = C with mica / kappa = 2/5nF = 0.4 nF

energy without mica = Q2/2C = 2002 / 2*0.4= 50000 nJ

So work to pull out mica = 50000-10000 = 40000 nJ

b) C without mica = C with mica / kappa = 2/5nF = 0.4 nF

Q does not change = CV = 2*100 = 200nC

Potential difference delta V = Q/C = 200/0.4 = 500 V

c)

energy with mica = Q2/2C = 2002 / 2*2 = 10000 nJ

energy without mica = Q2/2C = 2002 / 2*0.4= 50000 nJ

The change in energy is because the incuded dipole moment is removed when the mica is taken out.

d) If the capacitor remain connected to supply then voltage will remain constant instead of charge.

C without mica = C with mica / kappa = 2/5nF = 0.4 nF

Potential difference remains same, delta V = Q/C = 100 V

Q = CV = 0.4*100 = 40nC

energy with mica = Q2/2C = 2002 / 2*2 = 10000 nJ

energy without mica = Q2/2C = 402 / 2*0.4= 2000 nJ

Change in energy = 2000-10000 = -8000 J (so energy is released on removing mica as opposed to energy required to remove mica in the other case)

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