Parallel Plate Capacitor Connected to Battery: Answers

[shimizua]

Homework Statement

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 Volts. (C is the capacitance and U is the stored energy.) Give all correct answers concerning a parallel-plate capacitor charged by a battery (e.g. B, AC, CDF). If none of the statements is correct, enter the word none.

A) After being disconnected from the battery, inserting a dielectric with kappa > 1 will decrease V.
B) After being disconnected from the battery, increasing d increases U.
C) After being disconnected from the battery, increasing d decreases V.
D) With the capacitor connected to the battery, inserting a dielectric with kappa > 1 will increase U.
E) After being disconnected from the battery, inserting a dielectric with kappa > 1 will increase C.
F) After being disconnected from the battery, decreasing d decreases C.

The Attempt at a Solution

ok so i had thought it was ACD but that was said to be wrong. don't know where i am going wrong

 

[Delphi51]

I think C is false. Increasing d will decrease C, so V = Q/C will increase.
I would say D is false, too. Using V = Q/C in U=.5CV^2 = .5Q^2/C means that when C increases, U decreases.
E is true - the dielectric makes it a better capacitor with greater C
F is true - greater capacitance with reduced D.

 

[LowlyPion]

This should answer all your questions.
A good presentation of the material.

https://www.youtube.com/watch?v=E185G_JBd7U