A parallel-plate capacitor has plates with an area of 450 cm^2 and an air-filled gap between the plates that is 2.30 mm thick. The capacitor is charged by a battery to 570 V and then is disconnected from the battery.

How much energy is stored in the capacitor? (muJ)

The separation between the plates is now increased to 4.10 mm. How much energy is stored in the capacitor now?(muJ)

How much work is required to increase the separation of the plates from 2.30 mm to 4.10 mm? (muJ)

figure out C

E = (1/2) C V^2

figure new C and new E

work done = change in energy

To determine the energy stored in a capacitor, we use the formula:

E = (1/2) * C * V^2

where E is the energy stored, C is the capacitance, and V is the voltage.

To find the capacitance, we use the formula:

C = (ε₀ * A) / d

where C is the capacitance, ε₀ is the permittivity of free space (which is approximately 8.85 x 10^-12 F/m), A is the area of the plates, and d is the distance between the plates.

Let's calculate the energy stored in the capacitor using the given values.

1) Given:
Area of the plates, A = 450 cm^2 = 450 * 10^-4 m^2
Distance between the plates, d = 2.30 mm = 2.30 * 10^-3 m
Voltage, V = 570 V

First, calculate the capacitance:

C = (ε₀ * A) / d
C = (8.85 * 10^-12 F/m) * (450 * 10^-4 m^2) / (2.30 * 10^-3 m)
C = 7.94 * 10^-9 F

Now, calculate the energy stored:

E = (1/2) * C * V^2
E = (1/2) * (7.94 * 10^-9 F) * (570 V)^2
E ≈ 1.623 * 10^-3 J

Since the desired unit is microjoules (μJ), convert the answer to microjoules:

1 J = 1* 10^6 μJ
1.623 * 10^-3 J ≈ 1.623 * 10^-3 * 10^6 μJ
E ≈ 1623 μJ

Therefore, the initial energy stored in the capacitor is approximately 1623 μJ.

2) Now, let's calculate the energy stored when the separation between the plates is increased to 4.10 mm:

Given:
New distance between the plates, d' = 4.10 mm = 4.10 * 10^-3 m

Calculate the new capacitance using the same formula as before:

C' = (ε₀ * A) / d'
C' = (8.85 * 10^-12 F/m) * (450 * 10^-4 m^2) / (4.10 * 10^-3 m)
C' = 9.23 * 10^-10 F

Now, calculate the new energy stored:

E' = (1/2) * C' * V^2
E' = (1/2) * (9.23 * 10^-10 F) * (570 V)^2
E' ≈ 1.516 * 10^-3 J

Convert the answer to microjoules:

E' ≈ 1.516 * 10^-3 * 10^6 μJ
E' ≈ 1516 μJ

Therefore, the energy stored in the capacitor after increasing the separation to 4.10 mm is approximately 1516 μJ.

3) To calculate the work required to increase the separation of the plates from 2.30 mm to 4.10 mm, we use the formula:

Work = (1/2) * C * (V^2 - V'^2)

where V' is the voltage before the separation is increased.

Given:
Initial voltage, V = 570 V

Calculate the work:

Work = (1/2) * C * (V^2 - V'^2)
Work = (1/2) * (7.94 * 10^-9 F) * (570 V)^2 - (570 V')^2

To find V', we use the formula:

V' = V * sqrt(d' / d)

V' = 570 V * sqrt((4.10 * 10^-3) / (2.30 * 10^-3))
V' ≈ 794 V

Now, substitute the values into the work formula:

Work = (1/2) * (7.94 * 10^-9 F) * (570 V)^2 - (794 V)^2
Work ≈ 2.156 * 10^-4 J

Convert the answer to microjoules:

Work ≈ 2.156 * 10^-4 * 10^6 μJ
Work ≈ 215.6 μJ

Therefore, the work required to increase the separation of the plates from 2.30 mm to 4.10 mm is approximately 215.6 μJ.