An air filled parallel plate capacitor has plates of area 0.0406 m^2 and seperation of 2.5*10^-6 m.
a)What is the capacitence for the setup?
b)Our capacitor is connected to a 550 v battery, how much energy is stored in the capacitor?
c) If -1 mJ of work is performed by the capapcitor's plates but it is never disconnected from the 550 V battery, what is the new seperation of the plates?
a) C = epsilon*A/d
Epsilon = 8.85*10^-12 is the permittivity of free space.
I get C = 1.44*10^-7 F
b) Energy = (1/2) C V^2
I get 2.17*10^-2 J
c) If - 1 mJ is done BY the plates, then +1 mJ (0.001 J) is done ON the plates, by whatever force is changing the plate spacing. Electrical energy may also flow between capacitor and the battery while the plates are moved.
To get the work required, multiply the force between the plates by x = 0.001 m. The force is EQ*x were
E = V/d is the electric field and Q = CV is the charge.
Work = 0.001 J = (CV^2/d) * x
Solve for x and add it to the origial plate spacing. The plate spacing will be increased, but the stored energy in the capacitor will be less, because C will be reduced at the same voltage.
a) To calculate the capacitance of a parallel plate capacitor, you can use the formula:
C = ε₀ * A / d
Where:
C is the capacitance,
ε₀ is the permittivity of free space (approximately 8.85 × 10^-12 F/m),
A is the area of the plates, and
d is the separation between the plates.
Substituting the given values:
C = (8.85 × 10^-12 F/m) * 0.0406 m^2 / (2.5 × 10^-6 m)
C ≈ 5.71 × 10^-8 F
So, the capacitance of the setup is approximately 5.71 × 10^-8 Farads.
b) The energy stored in a capacitor can be calculated using the formula:
E = (1/2) * C * V^2
Where:
E is the energy stored in the capacitor,
C is the capacitance, and
V is the voltage across the capacitor.
Substituting the given values:
E = (1/2) * (5.71 × 10^-8 F) * (550 V)^2
E ≈ 8.90 × 10^-3 J
So, the energy stored in the capacitor is approximately 8.90 millijoules (mJ).
c) To find the new separation of the plates, we can use the work-energy principle. The work done on the capacitor plates is given by:
W = q * ΔV
Where:
W is the work done,
q is the charge on the capacitor plates, and
ΔV is the change in voltage across the capacitor.
But since the capacitor is connected to a battery, the charge q is given by:
q = C * ΔV
Substituting this expression for q in the work equation:
W = (C * ΔV) * ΔV
W = C * ΔV^2
Given that W = -1 mJ, we can set up the equation:
-1 × 10^-3 J = (5.71 × 10^-8 F) * ΔV^2
Solving for ΔV:
ΔV^2 = (-1 × 10^-3 J) / (5.71 × 10^-8 F)
ΔV ≈ ±313.75 V
Since the voltage cannot be negative in this scenario, we take the positive value:
ΔV ≈ 313.75 V
Now, we can use the formula for the parallel plate capacitor's capacitance to find the new separation d:
C = ε₀ * A / d
Rearranging the formula:
d = ε₀ * A / C
Substituting the given values:
d = (8.85 × 10^-12 F/m) * 0.0406 m^2 / (5.71 × 10^-8 F)
d ≈ 6.2 × 10^-5 m
So, the new separation of the plates is approximately 6.2 × 10^-5 meters.