two charges q1=-2uc and q2=3uc are placed at a distance of 6cm from each other. if rubber (Er=2.94) is placed between charges, find magnitude of electric force between charges.

To find the magnitude of the electric force between the charges when there is a material (in this case, rubber) with a relative permittivity between them, you can use the formula:

F = (1 / (4πε)) * ((q1 * q2) / r^2)

where:
F is the electric force,
π is a mathematical constant equal to approximately 3.14159,
ε is the permittivity of free space (also known as vacuum permittivity) and is approximately 8.854 × 10^-12 C^2/N·m^2,
q1 and q2 are the charges, and
r is the distance between the charges.

Given:
q1 = -2 μC (microCoulombs)
q2 = 3 μC (microCoulombs)
r = 6 cm = 0.06 m (converted to meters)
Er (relative permittivity of rubber) = 2.94

First, we need to adjust the permittivity value to account for the presence of the rubber.

The equation for the effective permittivity (Er-effective) is given by:

Er-effective = Er * ε

Now, substitute the given values:

Er-effective = 2.94 * 8.854 × 10^-12 C^2/N·m^2

Next, calculate the force:

F = (1 / (4π * Er-effective)) * ((q1 * q2) / r^2)

Substitute the values:

F = (1 / (4π * (2.94 * 8.854 × 10^-12))) * ((-2 * 10^-6) * (3 * 10^-6) / (0.06)^2)

Simplify and calculate:

F = (1 / (4π * (2.94 * 8.854 × 10^-12))) * ((-2 * 3) * 10^-12 / 0.0036)

F = (1 / (4π * (2.94 * 8.854 × 10^-12))) * (-6 * 10^-12 / 0.0036)

F = (-6 / (4π * (2.94 * 8.854 × 10^-12 * 0.0036))) * 10^-12

F = (-6 / (4π * (2.94 * 8.854 × 10^-12 * 0.0036))) * 10^-12

F ≈ -5.01 × 10^-2 N

Therefore, the magnitude of the electric force between the charges when there is rubber between them is approximately 5.01 × 10^-2 N. Note that the negative sign indicates that the force is attractive.