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.

 

[m2003]

Dear student,

Thank you for your questions about potential, capacitance, and charge in relation to capacitors. It is understandable to feel confused about these concepts at first, but with some clarification, you will have a better understanding of their relationships.

Firstly, it is important to understand that capacitance (C) is a measure of a capacitor's ability to store charge (Q) at a given potential difference (V). It is defined as the ratio of charge to potential difference (C = Q/V). This means that for a given capacitor, the capacitance is a fixed value and does not change.

Now, let's look at your first question. If the charge on the plates of a capacitor is doubled to 2Q, what happens to the capacitance and potential difference? The correct answer is (b) the capacitance becomes 2C. This is because the capacitance is a constant value and does not change, even when the charge is doubled. The potential difference also remains unchanged because the ratio of charge to potential difference (Q/V) stays the same.

For your second question, doubling the potential difference across a capacitor does not change its capacitance. This is because the capacitance is a constant value and does not depend on the potential difference.

Lastly, let's look at the effect of reducing the potential difference on the energy stored in a capacitor. The correct answer is (a) reduced to one-half. This is because the energy stored in a capacitor is directly proportional to the potential difference (U = 0.5*C*V^2). When the potential difference is reduced by half, the energy also reduces by half.

I hope this helps to clarify the relationships between potential, capacitance, and charge in capacitors. Remember, capacitance is a fixed value and does not change, and the ratio of charge to potential difference (Q/V) is also constant. Keep practicing and exploring these concepts, and you will have a better understanding in no time.

Best of luck with your studies!

Sincerely,