A voltage of 50 V appears across a 20-μF capacitor.

a) Determine the magnitude of the net charge stored on each plate.

b) Determinethe total net charge on both plates.

a. 1nC

b. 0C

Explanation:

Net charge q stored on plate of capacitor is

q = CV

Where C = 2uF = 2 x 10^-6F

V= 50v

q = 20 x 10^-6 x 50 = 1000 x 10^-6 = 1000uF = 10^-9 = 1nC

b. the total net charge on another plate is equal in absolute value to the first one, but it is charged with opposite Pole so always is valid that total net charge on both plates are equal to zero.

That's the other charge on the plate is -1nc

1 nC + -1nC = 0C

answer: 11/12

step-by-step explanation: 4 1/4 and 3 1/3 is equivalent to 4 3/12 and 3 4/12. 4 3/12 - 3 4/12 =11/12