To tackle your question, we’ll break it down into parts, addressing each aspect step by step. We’ll use Coulomb's Law to calculate the force between the charged spheres, analyze the nature of that force, and see how different materials affect it.

Calculating the Force Between Charged Spheres

First, let’s calculate the force between the two charged spheres using Coulomb's Law, which is given by the formula:

F = k * |q1 * q2| / r²

Where:

• F is the force between the charges.
• k is Coulomb's constant, approximately 8.99 x 10^9 N m²/C².
• q1 and q2 are the charges.
• r is the distance between the charges.

Given:

• q1 = 0.4 C
• q2 = 0.02 mC = 0.02 x 10^-3 C = 0.00002 C
• r = 30 cm = 0.3 m

Now, substituting the values into the formula:

F = (8.99 x 10^9 N m²/C²) * |0.4 C * 0.00002 C| / (0.3 m)²

Calculating the numerator:

0.4 C * 0.00002 C = 0.000008 C²

Now, substituting back:

F = (8.99 x 10^9) * 0.000008 / 0.09

F = (8.99 x 10^9) * 0.0000888889

F ≈ 799.2 N

Understanding the Nature of the Force

The force calculated is approximately 799.2 N. Since both charges are positive, the force between them is repulsive. This means that the two charged spheres will push away from each other due to their like charges.

Effect of Mica on the Force

Next, let’s consider the scenario where a sheet of mica, with a relative permittivity (dielectric constant) of 4, is placed between the spheres. The presence of a dielectric material reduces the electric field between the charges, which in turn affects the force.

The modified force can be calculated using the formula:

F' = F / εr

Where:

• F' is the new force.
• εr is the relative permittivity of the dielectric (4 for mica).

Substituting the values:

F' = 799.2 N / 4

F' = 199.8 N

Thus, the force between the spheres when mica is introduced is approximately 199.8 N, and it remains repulsive.

Introducing Iron Between the Charges

Now, if we introduce a sheet of iron, which is a conductor, between the charged spheres, the situation changes significantly. Conductors can shield electric fields. When a conductor is placed between two charges, it redistributes its charges in response to the electric field, effectively canceling the field in the region between the charges.

As a result, the force between the two charged spheres becomes negligible or effectively zero because the electric field is shielded by the conductor. Thus, the introduction of iron would eliminate the force between the charges.

Summary

To summarize:

• The initial force between the charged spheres is approximately 799.2 N and is repulsive.
• When mica is introduced, the force reduces to about 199.8 N, still repulsive.
• Introducing iron effectively cancels the force, bringing it to nearly zero.

Understanding these concepts helps illustrate how electric fields and materials interact in electrostatics. If you have any further questions or need clarification on any point, feel free to ask!