c=kQ/V right?
solve for c from that.
then
C=episilon*k*Area/separation right?
for thickness, you have breakdown voltage of 150Mv/m and that must equal 250/thickness. Solve for thickness. (if I were a design engineer, I would put a safety factor of at least 5 in that).
Now you have thickness (or separation), solve for area in the above equation.
On the discharge question, figure the initial charge (Q=CV/k) and the final charge, avg current equals charge that moved divided by time.
suppose you want to design a capacitor that uses mica as the dielectric material which has K = 7 and a dielectric strength of 150 MV/m. It is intended to store at least 1.- J of energy and be able to withstand a voltage up to 250 volts without electric breakdown.
a) What is the minimum thickness d of dielectric that can be used? For that thickness what is the area of a plate (and of the mica) that must be used?
b) Now assume the capacitor is initially charged to a voltage of 2.0 x 10^2 V. Connecting a wire from one plate to the other, the voltage falls to half its initial value in 2.4 ms. During that time, how large is the average electric current through the wire?
Any help would be appreciated. I was able to find the capacitance (3.2 x 10^-5 F) but do not know where to go after that. Thank you in advance!
1 answer