Problem &: A parallel-plate capacitor of capacitance Chas plate area A and dista

Question

Problem &: A parallel-plate capacitor of capacitance Chas plate area A and distance between plates d. The capacitor is concted to a battery with voltage V, fully charged and then disconnected. A slab of dielectric material with dielectric constant 4.0 is then inserted into capacitor,completely filling region between plates Grade Summary Feedback: 52% deduction 25% Part (c) After merting the delectrc, the darged stored in the capacitor is now: 25% Part (d) After mertingthe delectric, the cloctric field n the capacitor is now:

Explanation / Answer

Capacitance is given by:

C = e0*A/d

If we insert a dielectric, then capacitance will be given by:

C1 = k*e0*A/d = k*C

given that k = 4

C1 = 4*C

B.

We know that

Q = C*V

V = Q/C

after inserting dielectric, charge remains constant, So

V1 = Q/C1

V1 = Q/(4C) = (1/4)*(Q/C)

V1 = V/4

Part C.

We know that

Q = CV

Q1 = C1*V1

Q1 = 4*C*V/4 = C*V

Q1 = C*V

Part D.

Electric field is given by:

E = V/d

E1 = V1/d

E1 = (V/4)/d = (1/4)*(V/d)

E1 = E/4

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