The parallel plate of capacitor have each an area of 0.2m2 and are 10-2 m apart .The potential difference of 2000 V between them reduces to 1000 V. When asheet of dielectric material is inserted between plates ti fill the space between the plates completely Calculate 1 ) original capacitance of capacitors 2) capacitance after inserting dielectric material 3)charge on each plate 4) original electric field between plates 5) dielectric konstant of materal 6) electric field after inserting dielectric slab.?

To tackle this problem, we need to break it down into manageable parts. We will calculate the original capacitance, the capacitance after inserting the dielectric, the charge on each plate, the original electric field, the dielectric constant of the material, and the electric field after inserting the dielectric slab. Let's go through each step systematically.

1. Original Capacitance of the Capacitor

The formula for the capacitance (C) of a parallel plate capacitor is given by:

C = ε₀ * (A / d)

Where:

• ε₀ is the permittivity of free space, approximately 8.85 x 10^-12 F/m.
• A is the area of one of the plates (0.2 m²).
• d is the separation between the plates (0.01 m).

Plugging in the values:

C = 8.85 x 10^-12 F/m * (0.2 m² / 0.01 m)

C = 8.85 x 10^-12 F/m * 20

C = 1.77 x 10^-10 F or 177 pF.

2. Capacitance After Inserting Dielectric Material

The capacitance with a dielectric material is given by:

C' = κ * C

Where:

• C' is the new capacitance.
• κ is the dielectric constant of the material.

To find κ, we need to determine the new capacitance after inserting the dielectric. We can find it using the new voltage and charge relationship.

3. Charge on Each Plate

The charge (Q) on the capacitor can be calculated using the formula:

Q = C * V

Using the original capacitance and the original voltage:

Q = 1.77 x 10^-10 F * 2000 V

Q = 3.54 x 10^-7 C or 354 nC.

4. Original Electric Field Between Plates

The electric field (E) between the plates is calculated using:

E = V / d

Substituting the original voltage and distance:

E = 2000 V / 0.01 m

E = 200,000 V/m or 200 kV/m.

5. Dielectric Constant of the Material

After inserting the dielectric, the voltage reduces to 1000 V. The new capacitance can be calculated using the charge we found earlier:

C' = Q / V'

Where V' is the new voltage:

C' = 3.54 x 10^-7 C / 1000 V

C' = 3.54 x 10^-10 F or 354 pF.

Now, using the relationship between the original capacitance and the new capacitance:

κ = C' / C

κ = (3.54 x 10^-10 F) / (1.77 x 10^-10 F)

κ ≈ 2.

6. Electric Field After Inserting Dielectric Slab

Finally, we can calculate the electric field after inserting the dielectric:

E' = V' / d

Substituting the new voltage:

E' = 1000 V / 0.01 m

E' = 100,000 V/m or 100 kV/m.

Summary of Results

• Original Capacitance: 177 pF
• Capacitance with Dielectric: 354 pF
• Charge on Each Plate: 354 nC
• Original Electric Field: 200 kV/m
• Dielectric Constant: 2
• Electric Field with Dielectric: 100 kV/m

This systematic approach allows us to understand how capacitors behave with and without dielectric materials, and how these changes affect capacitance, charge, and electric fields. If you have any further questions or need clarification on any of these steps, feel free to ask!