We have enabled User registrations.
0 votes
edited by

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.

1 Answer

0 votes
selected by
Best answer

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

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.