Parallel plate capacitor problem

[smithnh]

[SOLVED] Capacitor problem

Problem:
A parallel plate capacitor with adjustible plate separation d and adjustible area A 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).

A) After being disconnected from the battery, increasing the area A will increase U.
B) After being disconnected from the battery, increasing d increases U.
C) After being disconnected from the battery, decreasing d increases C.
D) After being disconnected from the battery, increasing the area A will increase V.
E) With the capacitor connected to the battery, increasing the area A will increase C.
F) After being disconnected from the battery, increasing d decreases V.

Relevant equations:

(1) C=(epsilon_0)A/d

(2) V=E*d=(sigma_0)*d/(epsilon_0)

(3) U=(1/2)(epsilon_0)(E_field)^2*A*d

3. The Attempt at a Solution

A) seems to be true because of Eq (3)
B) seems to be true because of Eq (3)
C) seems to be true because of Eq (1)
D) seems to be false because the potential does not depend on the area in Eq (2). However, I also considered that the if the charge stays the same it will decrease the charge density, sigma_0 for increasing area and thus it will decrease V but this also supports the answer being false.
E) seems to be true because of Eq (1)
F) seems to be false because increasing d will increase V by Eq (2)

Answers already tried:
AB
ABE
ABCE
ABCDE

I cannot seem to figure this one out despite the fact that it seems so simple, where am I going wrong

 

[smithnh]

What I needed to think about was the fact that the battery cannot suplicate more charge thus Q is constant. After realizing that sigma_0 is defined as Q/A that then brought to question how I was thinking about the electric ield and was able to figure the answer.