What is the capacitance of the system?

[exitwound]

Homework Statement

Two metal objects have net charges of +70pC and -70pC, which result in a 20V potential difference between them.

a.) What is the capacitance of the system?
b.)If the charges are changed to +200pC and -200pC, what does the capacitance become?

Homework Equations

Q=CV
$C=\epsilon A/d$

The Attempt at a Solution

a.) Q=CV
C=Q/V = 70e-12 / 20 = 3.5pF

b.) I know the capacitance only relies on the area and the distance between the plates. But that is not obvious from Q=CV. Why can't this equation be used to find the answer by substituting in the 200pF for the charge and 20V? It's not obvious from these equations why and the book's descriptions absolutely suck.

 

[rl.bhat]

If you change the charges, potential difference will change.
If you remove the charges completely, still there will be the capacitor.

 

[exitwound]

Doesn't really help at all.

 

[rl.bhat]

Doesn't really help at all.

In the formula C = εA/d, there is no charge term of the potential difference term. The capacity of a capacitor independent of these quantities.

 

[mikelepore]

Why can't this equation be used to find the answer by substituting in the 200pF for the charge

Typing mistake noted. You meant to type 200 pC, not 200 pF.

and 20V?

The potential difference won't be 20 V anymore. Q=CV tells you that, when you change Q, you also change V.