Capacitor Problem, help!?
A 3.0 µF capacitor charged to 40 V and a 5.0 µF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other.
a. Find the final charge on each capacitor.
b. Find the potential difference across the capacitors.
Let Q = the charge on a capacitor in Coulombs
Let C = the capacitance in Farads
Let V = the voltage in volts
Q = CV
Q1 = (3.0 × 10^-6 F)(40 V) = 1.2 × 10^-4 C
Q2 = (5.0 × 10^-6 F)(18 V) = 9.0 × 10^-5 C
Because the capacitors are connected with the polarity reversed, (By the way, you should never to this, because lots of heat and flying metal may cause physical damage to anyone nearby.), we subtract the smaller from the larger:
Q1 – Q2 = 3 × 10^-5 C
The total capacitance is 8.0 × 10^-6 F
This will make the voltage:
V = 3.0 × 10^-5/8.0 × 10^-6 = 3.75 V