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

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm