Two charges `+ 10` and `+ 40` micro-coulomb are placed 12 cm apart in air. Find the position where electric field intensity is zero.

Explanation

Here , `q_1 = 10 xx 10^(-6) C, q_2 = 40 xx 10^(-6) C`
`AB = 12 cm = 0.12 m`
Let P be the point, where electric intensity due to `q_1` and `q_2` is equal and opposite i.e., net intensity is zero.
Let `AP = x, therefore BP = (0.12 - x)m`
Let a unit positive charge be placed at P.
Now, electric intensity at P due to `q_1` charge is
`vec(E_1) = 1/(4 pi in_0) (q_1)/(x^2)` along `vec(PB)`
`vec(E_1) = 9 xx 10^9 xx (10 xx 10^(-6))/(x^2) ` along `vec(PB)`
`vec(E_1) = (90 xx 10^(3))/(x^2)` along `vec(PB)`
and electric intensity at `P `due to `q_2` charge is
`vec(E_2) = 1/(4 pi in_0) (q_2)/((0.12 - x)^2) ` along `vec(PA)`
`vec(E_2) = (9 xx 10^9 xx 40 xx 10^(-6))/((0.12 - x^2)) = (360 xx 10^3)/((0.12 - x)^2)`
`E_1 = E_2` for zero intensity at P
or `1/(x^2) = 4/((0.12 - x)^2)`
Taking square roots on both sides
`1/x = pm 2/((0.12 - x))`
`0.12 - x = 2x` (taking + ve value)
or `3x = 0.12`
`x = 0.04 m`