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!
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.
To find the minimum thickness of the dielectric and the corresponding area of the plate, we can use the formula for capacitance of a parallel plate capacitor:
C = (ε * A) / d
where C is the capacitance, ε is the permittivity of the dielectric, A is the area of the plate, and d is the thickness of the dielectric.
a) Minimum thickness of the dielectric (d):
We will assume the capacitance is the minimum value required to store at least 1 J of energy, so C = 1 F. The permittivity (ε) can be calculated using the relative permittivity (K) of mica and the permittivity of free space (ε0) which is approximately 8.85 x 10^-12 F/m.
ε = K * ε0
= 7 * 8.85 x 10^-12 F/m
= 6.195 x 10^-11 F/m
Now, we can rearrange the formula to solve for d:
d = (ε * A) / C
Given that C = 1 F, ε = 6.195 x 10^-11 F/m, we can substitute these values in and solve for d.
b) Area of the plate for the minimum thickness (A):
Using the same formula, we can rearrange it to solve for A:
A = (C * d) / ε
Using the values C = 1 F, d from step a), and ε = 6.195 x 10^-11 F/m, we can substitute these values in and solve for A.
To find the average electric current during the discharge process, we can use the formula for current (I) in a capacitor:
I = (-dQ/dt)
where dQ is the change in charge and dt is the change in time.
During the discharge process, the voltage falls to half its initial value in 2.4 ms. We can assume the initial charge on the capacitor is Q0 = C * V0, where V0 is the initial voltage. The final voltage is Vf = V0/2.
The change in charge can be calculated as:
dQ = C * (Vf - V0)
Given the values of C from step a), V0 = 2.0 x 10^2 V, and Vf = V0/2, we can calculate dQ.
Finally, we divide dQ by dt (2.4 ms converted to seconds) to find the average electric current (I).