## In This post you will get answer of Q.

Two point charges 4$mu$C and – 2$mu$C are separated by a distance of 1 m in air. Then the distance of the point on the line joining the charges, where the resultant electric field is zero, is (in metre)

Solution:

## Let the point P where resultant field is zero be x m from 4$mu$C charge and (1- x) m distance apart from -2$mu$C charge. Since field is zero at this point then ,

$vec{E} , =vec{E_1}+vec{E_2} , = , 0$

$|vec{E}| , = , frac{1}{4piin_0}frac{frac{q_1}{2}}{r_1}+frac{1}{4piin_0}frac{frac{q_2}{2}}{r^2_2}$

$Rightarrow 0 =frac{1}{4pi in_0 }bigg[ frac{4mu C}{x^2}+frac{(-2mu C)}{(1-x^2)} bigg]$

$Rightarrow frac{4mu C}{x^2}=frac{2mu C}{(1-x^2)} Rightarrow frac{2}{x^2}=frac{1}{(1-x^2)}$

$Rightarrow 2(1-x)^2 , = , x^2$

Taking Root $sqrt{2}(1-x) , = , x$

$Rightarrow$ 1.414(1-x) , = , x $Rightarrow$ 1.414-1.414x , = , x

$Rightarrow$ 1.414 , = , (1+414)x $Rightarrow$x= $frac{1.414}{2.414}$

$Rightarrow$ x , = , 0.58m