? Two equal charges are placed in a medium at a distance of 9 cm from each other they exert a force equal to 4N on each other now one of the charge is increased by 6 microcoulomb so that the distance between them has to be increased by 3 cm in order to keep the force between then the same find the magnitude of charges and the dielectric constant of the medium?

Question

Answer

Given:

Two equal charges are placed in a medium at a distance of 9 cm from each other.

They exert a force equal to 4N on each other.

One of the charges is increased by 6 microcoulomb.

The distance between them has to be increased by 3 cm to keep the force between them the same.

To Find:

The magnitude of charges.

The dielectric constant of the medium.

Solution:

Step 1: Calculate the initial charges on each object.

Let the initial charge on each object be Q.

Using the Coulomb's law, we can find the magnitude of the charge Q:

$$F = \frac{kQ^2}{r^2}$$

where F = 4N, k = 9 x 10^9 Nm^2/C^2, and r = 9 cm = 0.09 m.

Solving for Q, we get:

$$Q^2 = \frac{Fr^2}{k}$$

$$Q^2 = \frac{(4)(0.09^2)}{9 imes 10^9}$$

$$Q = \sqrt{\frac{(4)(0.09^2)}{9 imes 10^9}}$$

$$Q = 2.4 imes 10^{-8} C$$

Therefore, each object has an initial charge of 2.4 x 10^-8 C.

Step 2: Calculate the final charge on the object that was increased.

Let the final charge on the object that was increased be Q1.

The other object still has a charge of Q.

Since the distance between them has increased by 3 cm to 12 cm, we can use Coulomb's law to find the new force:

$$F = \frac{kQ_1Q}{r^2}$$

where F = 4N, k = 9 x 10^9 Nm^2/C^2, and r = 12 cm = 0.12 m.

Solving for Q1, we get:

$$Q_1 = \frac{Fr^2}{kQ}$$

$$Q_1 = \frac{(4)(0.12^2)}{9 imes 10^9(2.4 imes 10^{-8})}$$

$$Q_1 = 2.0 imes 10^{-8} C$$

Therefore, the final charge on the object that was increased is 2.0 x 10^-8 C.

Step 3: Calculate the dielectric constant of the medium.

Since the force between the charges remains the same even after increasing the charge on one object and increasing the distance between them, we know that the medium is a dielectric.

To find the dielectric constant of the medium, we can use the formula:

$$k = \frac{1}{4\pi\epsilon_0}\frac{F r

Answer

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