Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is \(\frac{a}{{21}} \). × 10^-8C.The value of 'a' will be ______.

[Given g = 10 ms-2]

CONCEPT:

According to the coulomb's law, it is written as;

 F = k \(\frac{q_1q_2}{r^2}\)

Here we have, q₁ is the charge of the first mass, q₂ is the charge of the second mass, r is the distance and k is the proportionally constant.

CALCULATION:

Given: mass, m₁ = 10 mg

mass, m₂ = 10 mg

distance, d = 0.20 m

The free body diagram is written as;

With the help of a free body diagram we have;

T cosθ = mg = \(10 \times 10 ^{-6} \times 10\) = 10^-4     ----(1)

T sinθ = F_e 

Here, F_e = k \(\frac{q^2}{r^2}\)

Tsinθ =  k \(\frac{q^2}{d^2}\)

⇒ Tsinθ =  k \(\frac{q^2}{d^2}\) = F

⇒ Tsinθ =  k \(\frac{9 \times 10 ^9 \times q^2}{(0.2)^2}\) = F

⇒ Tsinθ =  k \(\frac{9 \times 10 ^9 \times q^2}{0.04}\) = F     ----(2)

Now, by dividing the equation (1) by (2) we have;

\(\frac{T sin θ }{T cosθ} = \frac {0.1}{\sqrt {0.24}}\)

⇒ tan\(θ\) = \( \frac {0.1}{\sqrt {0.24}} = \frac{F}{mg}\)    ----(3) 

Now, using equation (3) in equation (2) we have;

q = \(\frac{2\sqrt 10}{3\sqrt {\sqrt 24}} \times 10^{-8}\)

⇒ q = 0.95 \(\times 10^{-8}\)

⇒  q = 0.95 \(\times 10^{-8}\)

⇒ q =  \(\frac{20}{{21}} \). × 10^-8

Hence a = 20.