- #1

franktherabbit

- 12

- 1

## Homework Statement

This is a problem from some textbook a friend emailed me:

The text goes like this:

When the switch is open the charge on ##C1## equals ##Q_o=40uC## and ##C_2## has no charge on it. The switch closes and some charge ##q## flows through the capacitors (suppose the flow to be from up to down). Find:

a) ##q##

b) work of generator ##V##

c) the amount of work transferred into heat during this process

## Homework Equations

3. The Attempt at a Solution [/B]

Well the voltage when the switch is open exists only on the ##C1## capacitor. When it closes the amount of ##q## charge flows thorugh the capacitors in order to equalize the sum of voltages across the capacitors with V, is that right? That makes ##q=-10uC##, right? But why didnt ##R## have any role in here? Usually we have a voltage drop since we have resistors and here we do not?

For the work of the generator we have formula ##W=Vq## right? and that makes is ##-100uJ##.

Im not sure about the last part but i guess its just the difference of energies of the system?

##E_1=\frac{1}{2}(Q_o)^2/C_1=400uJ##

##E_2=\frac{1}{2}(Q_o+q)^C_1+\frac{1}{2}(q)^2/C_2+Eq=1/4(900)+1/4(100)-100=150uJ##

so the work transferred into heat is just ##E_2-E_1=-250uJ## right?