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 . × 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 

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^-4     ----(1)

T sinθ = F_e 

Here, F_e = k 

Tsinθ =  k 

⇒ Tsinθ =  k  = F

⇒ Tsinθ =  k  = F

⇒ Tsinθ =  k  = F     ----(2)

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

⇒ tan =     ----(3) 

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

q = 

⇒ q = 0.95 

⇒  q = 0.95 

⇒ q =  . × 10^-8

Hence a = 20.