A negative charge of 20 units is placed at a distance 50 cm away from a positive charge of 80 unit. Where will the electric field be zero on the line joining the two charges?

Explanation

Suppose, +80 unit and -20 unit of charges are place at A and B respectively [Fig 2.24]. The point where the electric field will be zero cannot lie in between A and B, because in that case intensity could be along the same direction, i.e., along `vec(AB)` for both the charges.
As the charge at A is greater, the point where the resultant intensity is zero will be situated on the right side of B, say at P. ltbr gt
Suppose, BP = x
Electric field at P due to the charge at B `=20/(x^(2)),` along `vec(PB)`
Electric field at P due to the charge at A =`80/((50+x)^(2)),` along `vec(AP)`
Since, the resultant intensity at P = 0
`therefore20/(x^(2))=80((50+x)^(2))` or, `x=50,-50/3`
Now, `xne-50/3` cm, because the point in that case would be situated in between A and B.
So, x = 50 cm, the polnt where the field is zero at a distance of 50 cm from the -20 unit charge on its right side.