Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is a × 10⁻⁸ C. The value of 'a' will be _______.

[Given g = 10 m/s⁻²]

To find the value of 'a' in the given problem, we can use the principles of electrostatics and mechanics. The two spheres are charged and repel each other, creating a balance between gravitational and electrostatic forces.

Given Data

• Mass of each sphere (m) = 10 mg = 10 × 10^-3 g = 10 × 10^-6 kg
• Length of thread (L) = 0.5 m
• Distance between spheres (d) = 0.20 m
• Acceleration due to gravity (g) = 10 m/s²

Forces Acting on the Spheres

The gravitational force (weight) acting on each sphere can be calculated as:

Weight (W) = m × g = (10 × 10^-6) × 10 = 10 × 10^-5 N

Electrostatic Force

The electrostatic force (F) between the two charged spheres can be expressed using Coulomb's law:

F = k × (q²) / d²

Where:

• k = 9 × 10⁹ N m²/C²
• q = charge on each sphere = a × 10^-8 C

Finding the Charge

In equilibrium, the vertical component of the tension in the thread balances the weight, while the horizontal component balances the electrostatic force. The geometry of the situation gives:

sin(θ) = d / (2L)

Here, θ is the angle between the thread and the vertical. We can find θ using:

sin(θ) = 0.20 / (2 × 0.5) = 0.20 / 1 = 0.20

Now, using the relationship between forces:

T × sin(θ) = F

And since T = W / cos(θ), we can substitute to find:

(W / cos(θ)) × sin(θ) = k × (q²) / d²

Calculating 'a'

Substituting the values and solving for 'a':

10 × 10^-5 × (0.20 / 1) = 9 × 10⁹ × ((a × 10^-8)²) / (0.20)²

After simplifying, we find:

a = 4.44

Thus, the value of 'a' is 4.44.