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)

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

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

where:
E is the energy stored in the capacitor,
C is the capacitance,
V is the voltage.

1. Find the capacitance (C):

C = (ε₀ * A) / d,

where:
ε₀ is the permittivity of free space (ε₀ = 8.85 * 10^-12 F/m),
A is the area of the plates (450 cm^2 = 0.045 m^2),
d is the separation between the plates (2.30 mm = 0.0023 m).

Plugging in the values:

C = (8.85 * 10^-12 F/m) * 0.045 m^2 / 0.0023 m.

C ≈ 1.73 * 10^-9 F.

2. Calculate the initial energy stored in the capacitor:

E₁ = (1/2) * C * V₁^2,

where:
V₁ is the voltage (V₁ = 570 V).

Plugging in the values:

E₁ = (1/2) * (1.73 * 10^-9 F) * (570 V)^2.

E₁ ≈ 0.289 J = 289 muJ.

3. Calculate the energy stored in the capacitor when the separation between the plates is increased to 4.10 mm:

Find the new capacitance (C₂):

C₂ = (ε₀ * A) / d₂,

where:
d₂ is the new separation between the plates (4.10 mm = 0.0041 m).

Plugging in the values:

C₂ = (8.85 * 10^-12 F/m) * 0.045 m^2 / 0.0041 m.

C₂ ≈ 9.62 * 10^-10 F.

Using the same formula as in step 2:

E₂ = (1/2) * C₂ * V₁^2,

where V₁ is the voltage (V₁ = 570 V).

Plugging in the values:

E₂ = (1/2) * (9.62 * 10^-10 F) * (570 V)^2.

E₂ ≈ 1.96 * 10^-4 J = 0.196 muJ.

4. Calculate the work required to increase the separation of the plates:

The work done in moving the plates apart is equal to the change in energy stored in the capacitor:

Work = E₂ - E₁.

Plugging in the values:

Work = (1.96 * 10^-4 J) - (0.289 J).

Work ≈ -0.289 J = -289 muJ.

Note: The negative sign indicates that work is done on the system (the capacitor) to increase the separation between the plates.

To find the energy stored in a capacitor, you can use the formula:

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

where C is the capacitance of the capacitor and V is the voltage across the capacitor.

To find the capacitance, you can use the formula:

C = (ε₀ * A) / d

where ε₀ is the permittivity of free space (which is a constant), A is the area of the plates, and d is the separation between the plates.

Let's calculate the energy stored in the capacitor with the given values:

1. For the initial state with a separation of 2.30 mm:

The area of the plates is 450 cm^2, which is equal to 450 * 10^-4 m^2 (since 1 cm^2 = 10^-4 m^2).

The separation between the plates is 2.30 mm, which is equal to 2.30 * 10^-3 m.

Using the formula for capacitance, we can calculate it:

C = (8.85 * 10^-12 F/m * (450 * 10^-4 m^2)) / (2.30 * 10^-3 m)
= 1.74 * 10^-10 F

Now, we can calculate the energy stored in the capacitor using the formula:

Energy = (1/2) * (1.74 * 10^-10 F * (570 V)^2)
= 0.070 Joules

To convert to microjoules (muJ), we multiply by 10^6:

Energy = 0.070 Joules * 10^6
= 70 muJ

Therefore, the initial energy stored in the capacitor is 70 muJ.

2. For the new state with a separation of 4.10 mm:

Using the same formula for capacitance:

C = (8.85 * 10^-12 F/m * (450 * 10^-4 m^2)) / (4.10 * 10^-3 m)
= 9.49 * 10^-11 F

Now, we can calculate the energy stored in the capacitor with the new capacitance and the same voltage:

Energy = (1/2) * (9.49 * 10^-11 F * (570 V)^2)
= 0.193 Joules

Converting to microjoules:

Energy = 0.193 Joules * 10^6
= 193 muJ

Therefore, the energy stored in the capacitor after doubling the separation is 193 muJ.

3. To find the work required to increase the separation from 2.30 mm to 4.10 mm, you can use the formula:

Work = (1/2) * (C * ΔV^2)

where ΔV is the change in voltage. Since the capacitor is disconnected from the battery, there is no current flowing, and the charge on the capacitor remains constant. Therefore, the voltage across the capacitor does not change.

Hence, the work required to increase the separation of the plates is zero.

Therefore, the work required to increase the separation from 2.30 mm to 4.10 mm is zero muJ.