Question

Two fixed charges

q

and

- 3 q

are placed on

x

-axis separated by distance

' d'

. Where should a third charge

2 q

be placed such that It will not experience any force?

Accepted Answer

If we place the third charge $2 q$ between $x$ and $' d'$ the direction of the force on $2 q$ due to $A$ and $B$ will be the same.So the net force cannot be zero.

If $2 q$ is placed the right side of $A$ then $r_{AC} > r_{BC} as q_{A} < q_{B}$ . So, $F_{CA} < F_{CB}$ always as the direction of $F_{CA}$ is towards right and $F_{CB}$ is left so $F_{CA} + F_{CB} \neq 0$ we can not obtain zero force.

Now consider $2 q$ at the $C$ to left of $q$ at distance $x$ from $q$ .
Force on $2 q$ at $C$ (left of $q$ ) is in opposite direction so net force will be zero if magnitude equal so,

$\Rightarrow F_{CA} + F_{CB} = 0$ or $F_{CA} = - F_{CB}$

$\Rightarrow \frac{KqcqA}{r^{2} CA} = \frac{- KqcqA}{r^{2} CB}$

$\Rightarrow \frac{2 q q}{x^{2}} = \frac{- 2 q (- 3 q)}{(x + d)^{2}}$

$\Rightarrow \frac{2 q^{2}}{x^{2}} = \frac{6 q^{2}}{(x + d)^{2}} \Rightarrow \frac{1}{x^{2}} = \frac{3}{(x + d)^{2}}$

$\Rightarrow 3 x^{2} = x^{2} + d^{2} + 2 xd$

$3 x^{2} - x^{2} - 2 xd - d^{2} = 0$

$2 x^{2} - 2 xd - d^{2} = 0$

$x = \frac{+ 2 d \pm \sqrt{(- 2 d)^{2} - 4.2 . (- d^{2})}}{2.2}$

$\Rightarrow x = \frac{+ 2 d \pm \sqrt{4 d^{2} + 8 d^{2}}}{4} = \frac{2 d \pm 2 d \sqrt{3}}{4}$

$\Rightarrow x = \frac{2 (d \pm d (\sqrt{3}))}{4} = \frac{d (1 \pm \sqrt{3})}{2}$

So, $x = \frac{d}{2} (1 + \sqrt{3}) m$ to the left of $q$ .

S.A

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