Two Particles A and B having charges q and 2q respectively are placed on a smooth table with a separation d. A third particle C is to be clamped on the table in such a way that the particles A and B remain at rest on the table under electrical forces. What should be the charge on C and where should it be clamped?
Given:
Charge of the particle A : q
Charge of the particle B : 2q
Distance between A and B : d
Formula used:
We will be using Coulomb’s Law:
Where ϵ0 is the permittivity of free space and it’s value is : ϵ0 = 8.85418782 × 10-12 m-3 kg-1 s4 A2 and F is the electrostatic force between two charges. And r is the distance between two charges.
The diagram below shows the given conditions:
C is the third particle clamped on the table. Let charge of point C be q’
For A and B to remain at rest, the electrostatic forces from A and B must cancel out each other. Which means magnitude of force from A to C : 
should be equal and opposite to the force from B to C : 
as Point B has twice the charge of A.
Thus the equation of rest is:
Here x is the distance between charge A and C.
From Coulomb’s Law,
Substituting we get,
Since we know that the magnitudes of FAC and FBC the forces are equal,
Rationalizing the denominator we get,
Substituting value of x in the equation (1) we get,
Hence, the charge on the C is -q(6-4√2) C and it should be clamped at a distance of (√2-1)d .