# Capacitor problem, easy but just can't seem to get it

• supermenscher

#### supermenscher

A 0.40E-6F and 0.70E-6F capacitor are connected in series to a 12 V battery. Calculate the potential difference and charge on each capacitor.

I know this problem is really easy, but I just can't seem to get it started. How do you find the potential difference. Can someone help me out?

supermenscher said:
A 0.40E-6F and 0.70E-6F capacitor are connected in series to a 12 V battery. Calculate the potential difference and charge on each capacitor.

I know this problem is really easy, but I just can't seem to get it started. How do you find the potential difference. Can someone help me out?
V = Q/C. You know the total voltage and total capacitance (I believe capacitors in series add up like resistors in parallel), and so you can calculate the total charge. Since they are in series, I would think that the charge on both capacitors would be the same, so you've got the charge values you need. Now, apply the V = Q/C equation again on each capacitor individually to get the voltage across each one, and you're done.

Thank you very much for your explanation, but it is still confusing me...do you think you could show me the first couple steps of what you mean to get me started...that would really help me out.

supermenscher said:
A 0.40E-6F and 0.70E-6F capacitor are connected in series to a 12 V battery. Calculate the potential difference and charge on each capacitor.

I know this problem is really easy, but I just can't seem to get it started. How do you find the potential difference. Can someone help me out?
Let $C_1$ and $C_2$ represent the capacitances of the capacitors. Let $C_t$ represent the capacitance of both capacitors in series.

$$C_t = (\frac{1}{C_1} + \frac{1}{C_2})^{-1}$$

$$V_t = \frac{Q_t}{C_t}$$

$$Q_t = V_tC_t$$

$$Q_1 = Q_2 = Q_t$$

$$V_1 = \frac{Q_1}{C_1}$$

$$V_2 = \frac{Q_2}{C_2}$$

The above is exactly what I said before, but in equation form. As long as I'm not mistaken, lines 1 and 4 are correct, and so you'll get the right answers. If I am mistaken, and one of them are wrong, you'll probably want to find the right equations somewhere for capacitance and charge in series.