Two point charges - q and ( q /2) are situated at the origin and at the point ( a , 0,0) respectively. The point along the x-axis where the electric field vanishes is

Question

Two point charges - q and ( q /2) are situated at the origin and at the point ( a , 0,0) respectively. The point along the x-axis where the electric field vanishes is (A) x=(a/√2) (B) x=√2 a (C) x=(√2 a/√2-1) (D) x=(√2 a/√2+1)

Tardigrade Admin · Accepted Answer

Suppose the field vanishes at a distance x we have ( Kq / x 2)=( Kq / 2/( x - a )2) 2( x - a )2= x 2 √2( x - a )= x √2 x-√2 a=x √2 x-x=√2 a x=(√2 a/√2-1) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/676e092f467945670cbf4f3c351fb3bb-.png" />

Q. Two point charges $- q$ and $\frac{q}{2}$ are situated at the origin and at the point $(a, 0, 0)$ respectively. The point along the x-axis where the electric field vanishes is

$x = \frac{a}{2}$

$x = 2$ a

$x = \frac{2 a}{2 - 1}$

$x = \frac{2 a}{2 + 1}$

Suppose the field vanishes at a distance $x$ we have
$\frac{K q}{x ^{2}} = \frac{K q /2}{( x - a ) ^{2}}$
$2 (x - a)^{2} = x^{2}$
$2 (x - a) = x$
$2 x - 2 a = x$
$2 x - x = 2 a$
$x = \frac{2 a}{2 - 1}$

Q. Two point charges −q and 2q​ are situated at the origin and at the point (a,0,0) respectively. The point along the x-axis where the electric field vanishes is

x=2​a​

x=2​ a

x=2​−12​a​

x=2​+12​a​