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.
Answer 1
Let Q = the charge of a capacitor in Coulombs
Let C = the capacity in Farads
Let V = the voltage in volts
Q = resume
Q1 = (3.0 × 10^-6F)(40V) = 1.2 × 10^-4C
Q2 = (5.0 × 10^-6F)(18V) = 9.0 × 10^-5C
Since the capacitors are connected with the polarity reversed (by the way, you should never do this, as too much heat and flying metal can cause physical damage to anyone nearby.), we subtract the smaller from the larger big :
Q1 – Q2 = 3 × 10^-5C
The total capacity is 8.0 × 10^-6 F
This will make the voltage:
V = 3.0 × 10^-5/8.0 × 10^-6 = 3.75V