A capacitor that has 10 cm diameter plates, 0.1mm apart, is charged at 100 V and the battery

removed afterwards. What are the charges on each of the platesimmediately after charging the
capacitor? What is the voltage if the distance between plates is increased to 0.5 mm? How
much did the energy stored in the capacitor change when plates are pulled apart?

To answer these questions, we need to understand the relationship between capacitance, charge, voltage, and separation distance in a capacitor.

1. What are the charges on each of the plates immediately after charging the capacitor?

The charge on each plate of a capacitor can be calculated using the formula:

Q = CV

Where Q is the charge, C is the capacitance, and V is the voltage.

To find the capacitance (C), we can use the formula:

C = εA/d

Where ε is the permittivity of the material between the plates, A is the area of the plates, and d is the separation distance between the plates.

Given:
- Diameter of the plates (D) = 10 cm = 0.1 m (since diameter = 2 * radius)
- Separation distance (d) = 0.1 mm = 0.1 * 10^(-3) m
- Voltage (V) = 100 V

First, let's calculate the area of the plates (A):

A = πr^2 (where r is the radius)

Since the diameter is given, we can calculate the radius:

r = D/2 = 0.1 m / 2 = 0.05 m

Now, substitute the values into the formula:

A = π * (0.05 m)^2

Now, calculate the capacitance (C):

C = εA/d

The permittivity (ε) depends on the material between the plates. Let's assume it is the permittivity of vacuum, which is approximately 8.85 x 10^(-12) F/m.

C = (8.85 x 10^(-12) F/m) * A / d

Substitute the calculated values:

C = (8.85 x 10^(-12) F/m) * (π * (0.05 m)^2) / (0.1 * 10^(-3) m)

Now, we have the capacitance. Substitute it into the formula for charge:

Q = CV

Q = [(8.85 x 10^(-12) F/m) * (π * (0.05 m)^2) / (0.1 * 10^(-3) m)] * 100 V

Calculate Q to get the charge on each plate.

2. What is the voltage if the distance between plates is increased to 0.5 mm?

To calculate the new voltage (V') when the separation distance (d') is increased to 0.5 mm, we need to use the formula:

V' = Q / C'

Where Q is the charge and C' is the updated capacitance.

Using the previous calculations, substitute the values:

V' = Q / [(8.85 x 10^(-12) F/m) * A / (0.5 * 10^(-3) m)]

Calculate V' to find the new voltage.

3. How much did the energy stored in the capacitor change when the plates are pulled apart?

The energy stored in a capacitor can be calculated using the formula:

E = 0.5 * C * V^2

To find the change in energy (ΔE), we need to compare the initial energy (E) with the energy when the plates are pulled apart (E').

Calculate the initial energy using the previous values:

E = 0.5 * C * V^2

Calculate the new energy (E') using the updated values:

E' = 0.5 * C' * V'^2

Now, calculate the change in energy by subtracting E from E'.

ΔE = E' - E

This will give us the change in energy stored in the capacitor.

By following these steps, you can find the charges on each plate immediately after charging the capacitor, the voltage if the separation distance is increased, and the change in energy stored in the capacitor when the plates are pulled apart.