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, select True or False.

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

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

After being disconnected from the battery, decreasing d increases U.

With the capacitor connected to the battery, inserting a dielectric with will decrease U.

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

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

Explanation / Answer

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

True

Since C = k*E*A/d, inserting a dielectric k will cause C to increase. Since charge Q is constant,

and Q = C*V, increasing C causes V to decrease

(b) With the capacitor connected to the battery, inserting a dielectric with will increase C.

True

Since C = k*E*A/d, inserting a dielectric k will cause C to increase

(c) After being disconnected from the battery, decreasing d increases U.

False

Charge Q is constant, energy stored is U = (0.5)*Q^2/C

Since C = E*A/d, decreasing d causes C to increase.

Since C increases, U then decreases

(d) With the capacitor connected to the battery, inserting a dielectric with will decrease U.

True

Charge Q is constant, energy store is U = (0.5)*Q^2/C

Since C = k*E*A/d, inserting dielectric k will cause C to increase.

Since C increases, U then decreases

(e) With the capacitor connected to the battery, decreasing d increases C.

True

Since C = E*A/d, decreasing d causes C to increases.

(f) After being disconnected from the battery, increasing d increases V.

True

Since C = E*A/d, increasing d causes C to decrease.

Since charge Q is constant, and Q = C*V, decreasing C causes V to increases.

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