# How do you find the charge and potential difference of a single capacitor? [closed]

Here is what the diagram of the capacitors look like:

  5.00 µF    6.00 µF
---| |--------| |---

---| |--------| |----
2.00 µF    4.00 µF


NOTE: The system of capacitors is connected to a 90 V Battery! And the Equivalent Capacitance of the entire system is 4.06 µF.

How do I find the charge on each of the 4 capacitors?
I have no idea how to approach this...I know I have to use couloumb's law but have no idea how to apply to each separate capacitor.

• The capacitors on the top part of the diagram are connected to the bottom part? – jim Jun 12 '16 at 8:52

You obviously know the formulae for finding the equivalent capacitance for series and parallel combinations.
Step back one pace to remember that capacitors in series have the same charge on them and capacitors in parallel have the same voltage across them.

So solve the problem by looking at the pairs of capacitors which are in series with one another not by looking at all four capacitors at once.

$$C = \frac {Q}{V}$$ $$Q = VC$$

You can use the capacitance ratios to find the voltages on them depending on how they're arranged with relation to the battery (it is unclear from your depiction). Then multiply the voltage by its capacitance to get the charge, $Q$.

• @M Barbosa Is it okay if you show me how? How would I calculate the charge on the 5.00 µF capacitor? I have no idea – HelloWorld Jun 12 '16 at 1:45
• We don't do solutions on here. Try to check out some tutorials on capacitor network analysis because this would be a pretty simple problem if you're familiar with that. Once you find the voltage across a capacitor, just multiply by its capacitance to find the charge. – M Barbosa Jun 12 '16 at 2:01

Hint: To find the effective capacitance you probably used the following steps:

(i) Replace the $5 \mu F$ and $6 \mu F$ capacitors by a single effective capacitor, say $C_1$;

(ii) Replace the $2 \mu F$ and $4 \mu F$ capacitors by another single effective capacitor, say $C_2$;

(iii) You now have two capacitors in parallel and you can now replace the original network by a single effective capacitor, say $C$.

(iv) Now retrace your steps: The capacitor in step (iii), $C$, has a 90 V battery connected across it. But this came from the workings in step (ii). This means that the p.d. across each of the terminals of each of the two (parallel) capacitors $C_1, C_2$ is 90 V so that you can calculate WHAT for each (effective) capacitor?

(v) Now go back one further step, each of the capacitors $C_1, C_2$ above is really a pair of capacitors in series. Just look at, for example, the top pair of series capacitors that you used to determine $C_1$. You know the charge on each of the capacitors since for capacitors in series are the same (as charge accumulates on the left hand plate of the $5 \mu F$ capacitor, an equal amount of charge is repelled on the right hand plate, this flows onto the left hand plate of the $6 \mu F$ plate, that repels an equal amount of charge on the right hand plate). This amount of charge was calculated in the step (iv).