Question

Calculate the value of two equal charges if they repel one another with a force of 1kgf when separated by a distance of 1cm in air. Now if the charges are situated in an insulating liquid having permittivity 8 times that of air then what would be the magnitude of the charges if they exert the same force on each other.​

Answer 1

Step-by-step explanation:

First, let's convert the force to SI units, since we'll need to use them in our calculation:

1 kgf = 9.81 N

The force between the charges is given by Coulomb's law:

F = k * q^2 / r^2

where F is the force, q is the charge, r is the distance between the charges, and k is Coulomb's constant:

k = 1 / (4πε0) = 8.99 x 10^9 N·m^2/C^2

where ε0 is the permittivity of free space.

We can use the given force and distance to solve for the value of q:

9.81 N = 8.99 x 10^9 N·m^2/C^2 * q^2 / (0.01 m)^2

Solving for q, we get:

q = ± 3.0 x 10^-6 C

The charges must have the same magnitude, so we take the positive value:

q = 3.0 x 10^-6 C

Now, if the charges are placed in a liquid with permittivity ε = 8ε0, the force between them will be:

F' = k * q^2 / (ε * r^2)

We want this force to be the same as before, so we can set it equal to 9.81 N and solve for q:

9.81 N = 8.99 x 10^9 N·m^2/C^2 * q^2 / (8ε0 * 0.01 m)^2

Solving for q, we get:

q = ± 6.0 x 10^-6 C

Again, we take the positive value since the charges have the same magnitude:

q = 6.0 x 10^-6 C

So the magnitude of the charges in the liquid is twice that in air.