A 0.55-μF capacitor is connected to a 3.5-V battery. How much charge is on each plate of the capacitor?

1.925μC

Explanation:

The relationship between charge (Q), voltage/potential difference (V) and capacitance (C) of a parallel plate capacitor is given by;

Q = CV      ----------------------(i)

Where, according to the question;

C = 0.55 μF = 0.55 x F

V = 3.5V

Substitute the values of C and V into equation (i);

=> Q = CV

=> Q = 0.55 x x 3.5

=> Q = 1.925 x C

=> Q = 1.925μC

Therefore the amount of charge on each plate of the capacitor is 1.925μC

it was given as a lesson and guide for prayer.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form