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

ID: 1575076 • Letter: A

Question

A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to 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 k will decrease U. increases U. decreases Q. dielectric with K will decrease C. increases C. dielectric with K will increase V. With the capacitor connected to the battery, increasing d With the capacitor connected to the battery, increasing d After being disconnected from the battery, inserting a With the capacitor connected to the battery, decreasing cd After being disconnected from the battery, inserting a Submit Answer Tries 0/20

Explanation / Answer

1) True

Increasing means C increases. Q is constant, so V = Q/C decreases.

U = ½CV² = ½QV = ½Q²/C Energy in a cap in Joules. Picking one with only one changing variable,

U = ½QV. Q is constant and V decreases, so energy decreases.

2) True

U = CV²/2 so energy will increase to U because capacitance will increase to C (V remains constant).

3) True

4) False

Capacitance will increase to C

5) True

6) False.

V will decrease. The dielectric sets up a back EMF which partially cancels V.