A distance of 2 m separates two point charges of `+5 xx 10^-19C`. Find the point on the line joining them at which electric field intensity is zero.

Explanation

Let the two charges `q_1 = -15 xx 10^(-19) C` and `q_2 = 5 xx 10^(-19) C` lie at A and B respectively.
Here, `r = AB = 2 m `
Let C is the point, where electric field is zero, at distance x from B.
So `BC = x `and `AC = r + x = 2 + x `
Let `E_1` = Electric field at C due to `q_1` at A.
`E_1 = 1/(4 pi in_0) (q_1)/(AC^2)` along CX
and `E_2` = Electric field at C due to `q_3` at B.
`E_2 = 1/(4pi in_0)(q_2)/(BC^2)` along CB
Electric field at C will be zero when `E_1 = E_2`
`implies 1/(4pi in_0) (q_1)/(AC^2) = 1/(4pi in_0)(q_2)/(BC^2)`
`implies (+ 15 xx 10^(-19))/((2 + x)^2) = (5 xx 10^(-19))/(x^2)`
`3/((2 + x)^2) = 1/(x^2)`
Taking square root on both sides
`implies (sqrt3)/((2 + x)) = 1/x`
`implies sqrt(3) x = 2 + x`
`implies sqrt(3)x - x = 2`
`implies x(sqrt(3) - 1) = 2`
`implies x (1.732 - 1) = 2`
` 0.732x = 2`
`x = 2/(0.732) = 2.73 m`
Electric field will be zero at distance `2.73 m` from `+5 xx 10^(-19)C` or at distance `(2 + 2.73 = 4.73m) 4.73 m` from charge `15 xx 10^(-19)C`.