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 C. After being disconnected from the battery, increasing d increases V. After being disconnected from the battery, inserting a dielectric with ? will increase U. After being disconnected from the battery, decreasing d decreases C. After being disconnected from the battery, decreasing d increases U. After being disconnected from the battery, inserting a dielectric with ? will increase V.

Explanation / Answer

With the capacitor connected to the battery, inserting a dielectric with will decrease C.
depends on k. if it is greater than one, as most all are, C will increase.
Parallel plate cap
C = r(A/d)
is 8.8542e-12 F/m
r is dielectric constant (vacuum = 1)
A and d are area of plate in m² and separation in m

With the capacitor connected to the battery, decreasing d increases U.
decreasing d increases the C
E = ½CV² Energy in a cap
Energy will increase

With the capacitor connected to the battery, decreasing d decreases Q.
Q = CV
decreasing d increases C, which increases Q

With the capacitor connected to the battery, inserting a dielectric with will decrease U.
new dielectric with k>1 will increase C, which increases Energy

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

After being disconnected from the battery, inserting a dielectric with will increase V.
new dielectric with k>1 will increase C
Q = CV
charge can't change, so
V = Q/C
C increases, so V decreases.

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