parallel plate capacitor has an area of 5cm^2 and a capacitance of 3.5PF. the capacitor is

connected to a 12v battery .after the capacitor is completely charge up the battery is removed.
a)what must be the separation of the plates and the charge density on the plate

To find the separation of the plates and the charge density on the plate, we can use the formula for the capacitance of a parallel plate capacitor:

C = (ϵ₀ * A) / d,

where C is the capacitance, ϵ₀ is the vacuum permittivity (a constant value), A is the area of the plates, and d is the separation between the plates.

Given:
Area (A) = 5 cm^2 = 5 * 10^(-4) m^2,
Capacitance (C) = 3.5 pF = 3.5 * 10^(-12) F.

Now, let's rearrange the equation to solve for d (the separation of the plates):

d = (ϵ₀ * A) / C,

where ϵ₀ is approximately 8.854 * 10^(-12) F/m (vacuum permittivity).

Substituting the given values into the equation:

d = (8.854 * 10^(-12) F/m) * (5 * 10^(-4) m^2) / (3.5 * 10^(-12) F).

Calculating this expression:

d ≈ 8.854 * 10^(-12) * 5 * 10^(-4) / 3.5 ≈ 1.261 * 10^(-6) m.

So, the separation of the plates is approximately 1.261 * 10^(-6) m.

To find the charge density (σ) on the plates, we can use the formula:

σ = Q / A,

where Q is the charge on the plates and A is the area of the plates.

When a parallel plate capacitor is charged by a battery, the charge on the plates can be calculated using the formula:

Q = C * V,

where V is the voltage applied (12 volts in this case).

Substituting the values:

Q = 3.5 * 10^(-12) F * 12 V ≈ 4.2 * 10^(-11) Coulombs.

Now, let's calculate the charge density:

σ = (4.2 * 10^(-11) C) / (5 * 10^(-4) m^2).

Calculating this expression:

σ ≈ 8.4 * 10^(-7) C/m^2.

So, the charge density on the plates is approximately 8.4 * 10^(-7) C/m^2.

To find the separation of the plates, we can use the formula for the capacitance of a parallel plate capacitor:

C = ε₀ * (A / d)

where C is the capacitance, ε₀ is the vacuum permittivity (8.85 x 10^-12 F/m), A is the area of the plates, and d is the separation of the plates.

Rearranging the formula, we get:

d = ε₀ * (A / C)

Using the given values:

A = 5 cm² = 5 x 10^-4 m²
C = 3.5 pF = 3.5 x 10^-12 F

Plugging the values into the formula, we get:

d = (8.85 x 10^-12 F/m) * (5 x 10^-4 m²) / (3.5 x 10^-12 F)

Simplifying the expression:

d = 1.26 x 10^-7 m = 126 nm

Therefore, the separation of the plates is 126 nm.

To find the charge density on the plates, we can use the formula:

Q = C * V

where Q is the charge on the plates, C is the capacitance, and V is the voltage.

Using the given values:

C = 3.5 x 10^-12 F
V = 12 V

Plugging the values into the formula, we get:

Q = (3.5 x 10^-12 F) * (12 V)

Simplifying the expression:

Q = 4.2 x 10^-11 C

The charge on the plates is 4.2 x 10^-11 C.

To find the charge density, we divide the charge by the area of one plate:

σ = Q / A

Using the given value:

A = 5 cm² = 5 x 10^-4 m²

Plugging the values into the formula, we get:

σ = (4.2 x 10^-11 C) / (5 x 10^-4 m²)

Simplifying the expression:

σ = 8.4 x 10^-7 C/m²

Therefore, the charge density on the plates is 8.4 x 10^-7 C/m².