Question:

A parallel plate capacitor consists of two square plates with side lengths of 10 cm each and a separation distance of 2.0 mm. The plate on the left is connected to a 12 V battery, while the plate on the right is connected to the ground.

a) Calculate the capacitance of this parallel plate capacitor.

b) If a dielectric material with a dielectric constant of 5.0 is inserted between the plates, calculate the new capacitance.

c) Calculate the charge stored on each plate of the capacitor when it is fully charged.

d) Calculate the electric field between the plates of the capacitor.

Answer:

a) The capacitance of a parallel plate capacitor can be calculated using the formula:

Where:

C is the capacitance,

A is the area of one plate,

ε₀ is the permittivity of free space, and

d is the separation distance between the plates.

Given: A = (10 cm)² = (0.10 m)² = 0.01 m² d = 2.0 mm = 0.002 m ε₀ = 8.85 × 10⁻¹² F/m (permittivity of free space)

Substituting the values into the formula, we can find the capacitance:

Calculating further:

C = 0.01 F

Therefore, the capacitance of the parallel plate capacitor is 0.01 F.

b) When a dielectric material is inserted between the plates, the capacitance increases by a factor equal to the dielectric constant (K) of the material.

Given: K = 5.0

The new capacitance (C') can be calculated by multiplying the original capacitance (C) by the dielectric constant (K):

Substituting the values:

C' = 5.0 × 0.01 F = 0.05 F

Therefore, the new capacitance after the dielectric material is inserted is 0.05 F.

c) The charge stored on each plate of the capacitor when it is fully charged can be calculated using the formula:

Where:

Q is the charge stored on each plate, and

V is the voltage applied to the capacitor.

Given: V = 12 V

Substituting the values:

Q = 0.01 F × 12 V = 0.12 C

Therefore, the charge stored on each plate of the capacitor when it is fully charged is 0.12 C.

d) The electric field between the plates of the capacitor can be calculated using the formula:

Where:

E is the electric field,

V is the voltage applied to the capacitor, and

d is the separation distance between the plates.

Given: V = 12 V d = 2.0 mm = 0.002 m

Substituting the values:

E = 12 V / 0.002 m = 6,000 N/C

Therefore, the electric field between the plates of the capacitor is 6,000 N/C.