How Does Doubling Charge or Potential Affect a Capacitor's Behavior?

[Soaring Crane]

I’m a bit confused on the relationships among C,V, and Q after reading my text. I don’t know if my reasonings contradict each other for the following.

Homework Statement

A capacitor of capacitance C holds a charge Q when the potential difference across the plates is V. If the charge Q on the plates is doubled to 2Q,

a. the capacitance becomes (1/2)V.
b. the capacitance becomes 2C.
c. the potential difference changes to (1/2)V.
d. the potential difference changes to 2V.
e. the potential difference remains unchanged.

Homework Equations

C = Q/V

The Attempt at a Solution

The ratio Q/V is a constant, and Q is directly proportional to V. Therefore, if the charge is doubled, won’t the V, the potential difference, double also?

Homework Statement

Doubling the potential difference across a capacitor

a. doubles its capacitance.
b. halves its capacitance.
c. quadruples the charge stored on the capacitor.
d. halves the charge stored on the capacitor.
e. does not change the capacitance of the capacitor.

Homework Equations

C = Q/V or C = (epsilon_0*A)/(d)

The Attempt at a Solution

Following the fact that Q is proportional to V and the ratio of Q to V does not change, the capacitance does not change?

Homework Statement

If the potential difference of a capacitor is reduced by one-half, the energy stored in that capacitor is

a. reduced to one-half.
b. reduced to one-quarter.
c. increased by a factor of 2.
d. increased by a factor of 4.
e. not changed

Homework Equations

U = (0.5)*C*V^2

The Attempt at a Solution

U = (0.5)*C*V^2

U2 = (0.5)*C*(0.5*V)^2 = 0.5*C*[(V^2)/4] = (1/2)*(1/4)*C*V^2 = (1/4)*U

The energy decreases by one-quarter?

Thanks.

 

[berkeman]

Correct, correct, and correct. You don't seem confused to me. Changing the capacitance requires a physical change to the shape or size of the cap, not just Q or V changes.