Question

Two small balls having equal positive charge Q each are suspended by two insulating strings of equal length L, from a hook fixed to a stand. The whole set up is taken in
a satellite into space. The tension in each string is
Option:
a) 0
b) 1/4πε₀ × Q²/L²
c) 1/4πε₀ × Q²/2L²
d) 1/4πε₀ × Q²/4L²

Answer 1

Answer:

Option d) 1/4πε₀ × Q²/4L²

Explanation:

According to coulomb’s law, the magnitude of electrostatic force between the charges is given by,

F ∝ q¹q²r²

F=kq¹q²r²…………. (1)

Where k is a constant of proportionality called electrostatic force constant. F is the force, q¹q² charges, d is the distance

We have k=1/4πε₀

Substituting the value of k is equation 1 we get,

From equation (1), 1/4πε₀×q¹q²r² ……………(2)

Let us consider two positive charges +Q is suspended by a two insulating string and charges are separated.

By a distance L from the fixed-point O. Now we can observe the force of repulsion between the charges and they are trying to move away from each other by applying force F.

Now the whole set up is taken in a satellite into space where there is no gravity (state of weightlessness), then mg=0

In the absence of gravitational force, they stand horizontal and opposite to each other at an angle 180∘.

Therefore, the angle between the two strings is 180∘.

The Force of repulsion between the positive charges is F, which equal to the tension in the string.

Thus electrostatic force=tension in the string.

Since there is no gravity.

We are aware of the coulomb’s law; it states that the electrostatic force between two stationary point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between their centers.

Electrostatic force, F=1/4πε₀×q¹q²/r²

Then, using the equation, where, q¹=Q, q²=Q and r=2L= distance between two charges.

We get, F=T ⇒ 1/4πε₀ × Q²/(2L)²

∴ Tension in each string T= 1/4πε₀ × Q²/4L²