# Capacitors in Series

#### runfast220

The drawing shows two fully charged capacitors(C1= 2.00 microF, q1=6.00 microC; C2=8.00 microF, q2=12.0 microC). The switch is closed, and the charge flows until equlibrium is reestablished(i.e until both capacitors have the same voltage across their plates). Find the resulting voltage across either capacitor.

I lost at what to do. The first thing I would do would to be find the V for each capacitor V=q/C, but after that I'm clueless. The answer in the book says 1.8V.

#### arze

i don't see any drawing.

#### runfast220

Sorry...here is the diagram.

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#### arze

When the switch is closed, the charge moves toward equilibrium in the circuit. The total charge is 12+6=18mC, and the toal capacitance is2+8=10mF. Voltage across the two capacitors is equal. So $$\displaystyle V=\frac{q}{C}=\frac{18}{10}=10V$$

runfast220

#### Pmb

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When the switch is closed, the charge moves toward equilibrium in the circuit. The total charge is 12+6=18mC, and the toal capacitance is2+8=10mF. Voltage across the two capacitors is equal. So $$\displaystyle V=\frac{q}{C}=\frac{18}{10}=10V$$
Notice that the answer runfast220 quoted was V = 1.8 Volts. Your method is correct but you made a simple arithmetic error. The answer is

$$\displaystyle V=\frac{Q_{total}}{C_{total}}=\frac{18\mu F}{10\mu C} = 1.8 V$$

I love math. I hate arithmetic.

arze

#### arze

oops! that was a typo. haha