Two identical metallic spheres having unequal opposite charges are placed at a distance 0.9 m apart in air. After bringing them in contact with each other they are again placed at the same distance apart. Now the force of repulsion between them is 0.025 N. Calculate the final charges on each of them.

To find the final charges on the two metallic spheres after they have been brought into contact and then separated, we can follow these steps:

Initial Setup

Let the initial charges on the spheres be +Q and -q. When they come into contact, the total charge is shared equally because they are identical spheres.

Total Charge Calculation

The total charge before contact is:

• Total Charge = +Q - q = Q - q

After contact, the charge on each sphere becomes:

• Charge on each sphere = (Q - q) / 2

Force of Repulsion

After being separated, the force of repulsion between the spheres is given by Coulomb's law:

• F = k * |q1 * q2| / r²

Here, F is the force (0.025 N), r is the distance (0.9 m), and k is Coulomb's constant (approximately 8.99 x 10⁹ N m²/C²).

Substituting Values

Let the final charges on the spheres be q1 and q2, where:

• q1 = (Q - q) / 2
• q2 = (Q - q) / 2

Since they are equal after contact, we can denote both as q.

Setting Up the Equation

Now, substituting into Coulomb's law:

• 0.025 = (8.99 x 10⁹) * (q * q) / (0.9)²

Solving for Charge

Rearranging gives:

• q² = (0.025 * (0.9)²) / (8.99 x 10⁹)

Calculating this will yield:

• q² ≈ 2.25 x 10^-12
• q ≈ 1.5 x 10^-6 C

Final Charges

Thus, the final charges on each sphere after contact and separation are:

• Charge on Sphere 1: +1.5 x 10^-6 C
• Charge on Sphere 2: -1.5 x 10^-6 C