A parallel plate capacitor with plate separation d is connected to a battery. Th

Question

A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V. (C is the capacitance and U is the stored energy.) Answer the following questions regarding the capacitor charged by a battery. For each statement below,True or False.

1. With the capacitor connected to the battery, decreasing d increases C.

2. After being disconnected from the battery, increasing d increases V.

3. After being disconnected from the battery, inserting a dielectric with will decrease V.

4. After being disconnected from the battery, inserting a dielectric with will decrease U.

5. With the capacitor connected to the battery, increasing d increases U.

6. With the capacitor connected to the battery, inserting a dielectric with will increase C.

Explanation / Answer

C = eoA/d

Q = C*V

U = 0.5*C*V^2 = 0.5*Q*V = 0.5*Q^2/C

(a)

here V is constant

C = eo*A/d

as d is decreased C increases

TRUE

---------

2)

here the Q remains same
Q = C*V

Q1 = Q2

C1*V1 = C2*V2

as d increases the C decreases

C2 < C1

V2 = V1*(C1/C2)

V2 > V1

V increases

TRUE

+++++

3)

Q1 = Q2

C1*V1 = C2*V2

as the dielectric is inserted , the capacitance increases

C2 > C1

V2 = V1*(C1/C2)

v2 < V1

V decreases

TRUE

--------------

4)

U1 = 0.5*Q*V1

U2 =0.5*Q*V2

V2 < V1

U2 < U1

energy decreases

TRUE

++++++++++++++++

5)

V is constant

as d increases C decrease

U = 0.5*C*V^2

U also decreases

TRUE

+++++++++++++++++++++

C = eo*A/d

dielectric is inserted

C' = k*eo*A/d = k*C

capacitance increases

TRUE

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