Question

Two charges q and –3q are placed fixed on x-axis separated by distance ‘d'. where should a third charge 2q be placed such that it will not experience any force

Answer 1

Hope it helps you out with the problem

Answer 2

Hey!!

Here is your answer -

Following Coulomb's Law:

F = k×q1×q2/d^2

Where,  

q1 is charge 1

q2 is charge 2

k is a constant

And d is the distance between the 2 charges

We want the forces to be equal, so F1 =F2

F1= k×q1×q2/d1^2

F2 = k×q3×q2/d2^2

The k and q2 terms cancel out, we end up with:

q1/d1^2 = q3/d2^2

Take both d to one side and the q to another, So the ratio of the distances will be:

d1/d2= sqrt(3)/3 (notice that I removed the minus sign because it only means that q3 and q2 attract)

Now to the final solution.

We know they all lie on x- axis and that q1 and q2 will repel and q3 and q2 will attract.

This means that q1 is between q2 and q3, to ensure zero force on q2. (Because if q2 is between q1 and q3, the force will be to the direction of q3 since q3 pulls it and q1 pushes it away).

So lets assume q1 location is our reference point, the distance between q1 and q2 is d1

And we know that the distance between q1 and q3 is d (given in the question)  

This means d2 = d1 +d

So our equation will be

d1/(d1+d) = 0.577 (roughly)

So d1 = 0.577×d1 + 0.577×d

d1 - 0.577×d1=0.577×d

0.423×d1=0.577×d

d1= 1.364×d