Question

2 point charges +2q and +4q are seperated by a distance of 6a. Find the point on the line joining 2 charges where electric field is zero.​

Answer 1

Two-point charges +2q and +4q are separated by a distance of 6a.

Let the first charge i.e. +2q is placed at point A and second charge B i.e. +4q is placed at B point and the distance between both the charges is 6a.

We have to find the point on the line joining 2 charges where the electric field is zero.

E(vector) = F(vector)/q

F = k (q1q2)/r²

Assume that a charge +Q is placed between point A and point B. Also, +Q is placed at a distance of x from the A point.

Now,

F = k (2q)/x²

Also,

F = k (4q)/(6a - x)²

On comparing we get,

k (2q)/x² = k (4q)/(6a - x)²

2q/x²= 4q/(6a - x)²

1/x² = 2/(6a - x)²

Cross-multiply them

2x² = (6a - x)²

√2 x = 6a - x

√2x + x = 6a

x(√2 + 1) = 6a

x = 6a/(√2 + 1)

Rationalize it

x = 6a(√2 - 1)/(2 - 1)

x = 6a(√2 - 1)/1

x = 6a(√2 - 1)