A 1.50-µF capacitor charged to 40.0 V and a 2.40-µF capacitor charged to 15.0 V are connected to each other, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 1.50-µF capacitor?

Question

anonymous · Answer

@Esraa ,  can you draw the circuit diagram of this?

anonymous · Answer

Are they connected in series or parallel?

esraa · Answer

I've no idea this is the question :(

anonymous · Answer

This is parallel
Capacitance divides.

anonymous · Answer

The final charge would be the max charge of the cap, given no time of run, and no max V on either cap

anonymous · Answer

They are connected something like this.
|dw:1335799551640:dw|
When they are connected in this manner, what will happen is that the negative charge on each plate removes the positive charge on the other plate. The total charge will decrease as a result. The remaining charge will be distributed in a manner such that the potential difference between both capacitors is equal.
Calculate this total charge left.
$$V _{1}C _{1}-V _{2}C _{2}$$
This is the charge that is distributed among the two capacitors such that potential difference is same across plates. 
You should be able to show that for this to happen, the charges should be distributed in the ratio of their capacitance.

esraa · Answer

@marwa.ally  momken teshoofy da?

anonymous · Answer

el diagram fen ??

esraa · Answer

mafeesh diagram

anonymous · Answer

Ah yes, that's right.
The equation should be enough... Isn't it?

esraa · Answer

Yeah it should be

esraa · Answer

@ash2326