> 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) – LIVE ANSWER TODAY

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)

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

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