Help! Calculating Capacitance from Voltage & Charge

[nw0rbrolyat]

Capacitance (Please Help!)

Homework Statement

The charge on a capacitor increases by 21 \mu(micro) C when the voltage across it increases from 99 V to 127 V.

What is the capacitance of the capacitor?

Homework Equations

There is Q=cV, or C=Q/v. But there are 2 unknowns and no number for the charge? I am confused!

The Attempt at a Solution

I don't even know where to start? =/
This is all that was given in the question.

 

[nasu]

1. The problem statement, all variables and given/known
There is Q=cV, or C=Q/v. But there are 2 unknowns and no number for the charge? I am confused!

There are also two equations, so it should be OK.
Write the equation for each state (initial V=99 V and final V=127 V) and subtract the two equations.

 

[nw0rbrolyat]

I don't think I am following you. Use which equation?
I did x(99)=(x+21)(127) and got -95.25
I used C=QV

 

[nasu]

C=QV is incorrect.
I mean the equation Q=CV

 

[Xerxes1986]

All you know is that charge increased when you changed your voltage, which should be apparent from the equation Q=CV. This is a linear equation meaning that if you graphed Q as a function of V you would get a straight line whose slope is C. If you recall from geometry,

$$m=\frac{\Delta Y}{\Delta X}$$

So dividing your change in Q by your change in V should yield C.

Alternatively, you can solve the system of 2 equations and 2 unknowns:

$$Q_{i}=C V_{i}$$
$$Q_{f}=C V_{f}$$