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## 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.