Solve Unknown Capacitance from Series Capacitors & Battery Charge

[Angie K.]

Homework Statement

A 197-pF capacitor is connected in series with an unknown capacitor, and as a series combination they are connected to a 25.4-V battery. If the 197-pF capacitor stores 143 pC of charge on its plates, what is the unknown capacitance?

Homework Equations

Q = C*V (charge stored on each capacitor)
Total Capacitance = 1/C1+1/C2
Total Charge Stored = QTotal = CTotal*VTotal

The Attempt at a Solution

I don't even know where to start. It seems like I should just be able to use one of the equations above but just rearrange it so I am solving for the unknown capacitor.

 

[Angie K.]

I know that the charge of capacitors in series is equal.

 

[Angie K.]

Homework Statement

A 197-pF capacitor is connected in series with an unknown capacitor, and as a series combination they are connected to a 25.4-V battery. If the 197-pF capacitor stores 143 pC of charge on its plates, what is the unknown capacitance?

Homework Equations

Q = C*V (charge stored on each capacitor)
Total Capacitance = 1/C1+1/C2
Total Charge Stored = QTotal = CTotal*VTotal

The Attempt at a Solution

I tried using the equation V(total)=Q(total)/C(total)
Known variables plugged in:
25.4V = 143pC/(197+x)
and I tried to solve for x (the second unknown capacitor) using algebra.
But that didn't work because for x, I got a value of -191.37pF

 

[gneill]

Capacitors in series don't add like C1 + C2. That's why your attempt at a solution went awry and you came up with a negative value.

But your observation that the charge on capacitors in series are equal is valid, and will prove to be useful here.

Since you're given the capacitance and charge on one of the capacitors you should be able to determine the voltage across that capacitor. What then is the voltage on the other capacitor? What's the charge? What then is its capacitance?