Question

two charges q and minus 3 q are placed fixed on x-axis separated by a distance D where should a third charge first it will be placed such that it will not experience any force

Answer 1

in the above pic instead of D-x its D+x and the answer will be D/0.73 and pls verify it as its answer may not match

Answer 2

Explanation:

At P: on 2q, Force due to -3q is to the right and q is to the left.

$\frac{2 q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{6 q^{2}}{4 \pi \varepsilon_{0}(d+x)^{2}}$

On cancelling, we get,

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

$\Rightarrow d^{2}+x^{2}+2 x d-3 x^{2}=0$

On rearranging, we get,

$\Rightarrow d^{2}-2 x^{2}+2 x d=0$

$\Rightarrow 2 x^{2}-2 x d-d^{2}=0$

To find roots, we know that,

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Where, a = 2; b = -2d; c = $-d^{2}$

$x=\frac{2 d \pm \sqrt{4 d^{2}-\left(4 \times 2 \times-d^{2}\right)}}{4}$

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

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

$x=\frac{2\left(d \pm \sqrt{3 d^{2}}\right)}{4}$

$x=\frac{d \pm \sqrt{3 d^{2}}}{2}$

$x=\frac{d \pm \sqrt{3} d}{2}$

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

To the left of q.

Negative sign lies between q and -3q and hence is unadaptable.

We have used the concept of electrostatics in this question to calculate the position of a third charge such that it will not experience any force.

$x=\frac{d}{2}(1+\sqrt{3})$