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

See prev post

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

C = (ε₀ * A) / d,

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

Given:
A = 5 cm^2 = 5 x 10^-4 m^2,
C = 3.5 pF = 3.5 x 10^-12 F.

First, let's find the separation of the plates (d):

Rearranging the formula: d = (ε₀ * A) / C

Substituting the given values:
d = (8.85 x 10^-12 F/m) * (5 x 10^-4 m^2) / (3.5 x 10^-12 F)

Now, calculate the value of d:

d = (8.85 x 5 x 10^-16) / (3.5 x 10^-12) = 1.2714 x 10^-4 m

So, the separation of the plates is approximately 1.2714 x 10^-4 m.

Next, let's find the charge density (σ) on the plates. The charge density is the total charge (Q) divided by the area (A):

σ = Q / A

We can calculate Q by using the formula:

Q = C * V,

where:
Q is the charge on the capacitor,
C is the capacitance,
V is the applied voltage.

Given:
C = 3.5 x 10^-12 F,
V = 12 V.

Calculating Q:

Q = (3.5 x 10^-12 F) * (12 V) = 4.2 x 10^-11 C

Now, calculate the charge density:

σ = (4.2 x 10^-11 C) / (5 x 10^-4 m^2) = 8.4 x 10^-7 C/m^2

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