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!
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To design a capacitor that uses mica as the dielectric material, we need to consider the specifications given: a dielectric constant (K) of 7, a dielectric strength of 150 MV/m, a minimum energy storage of 1 J, and a breakdown voltage of 250 V. Let's break down the problem into two parts:
a) Minimum thickness of the dielectric and corresponding plate area:
The formula for the capacitance of a parallel plate capacitor is:
C = (ε₀ * K * A) / d
where C is the capacitance, ε₀ is the vacuum permittivity (8.854 x 10^-12 F/m), K is the dielectric constant, A is the area of one plate, and d is the thickness of the dielectric.
We have the capacitance, C = 3.2 x 10^-5 F, and we want to find the minimum thickness (d) and its corresponding plate area (A).
Rearranging the formula, we get:
d = (ε₀ * K * A) / C
Substituting the known values:
d = (8.854 x 10^-12 F/m * 7 * A) / (3.2 x 10^-5 F)
d = 1.944 x 10^-6 A (in meters)
Next, we need to find the plate area (A):
Area can be calculated using the formula:
A = C * d / (ε₀ * K)
Substituting the known values:
A = (3.2 x 10^-5 F * 1.944 x 10^-6 m) / (8.854 x 10^-12 F/m * 7)
A = 0.8862 m²
Therefore, the minimum thickness of the dielectric is approximately 1.944 μm, and the corresponding plate area is approximately 0.8862 m².
b) Average electric current through the wire:
To find the average electric current through the wire, we need to use the formula:
I = ΔQ / Δt
where I is the average current, ΔQ is the change in charge, and Δt is the time interval.
We are given that the voltage dropped to half its initial value in 2.4 ms, or 2.4 x 10^-3 s.
The change in charge (ΔQ) is given by:
ΔQ = C * (V_final - V_initial)
Substituting the known values:
ΔQ = (3.2 x 10^-5 F) * (0.5 * 2.0 x 10^2 V - 2.0 x 10^2 V)
ΔQ = -3.2 x 10^-5 C
Since the charge cannot be negative, we take the magnitude of ΔQ:
ΔQ = 3.2 x 10^-5 C
Now, substituting the values into the formula for average current:
I = (3.2 x 10^-5 C) / (2.4 x 10^-3 s)
I ≈ 0.0133 A or 13.3 mA (rounded to three significant figures)
Therefore, the average electric current through the wire is approximately 0.0133 Amperes or 13.3 milliamperes.