If the net electric field at point P along Y axis is zero, then the ratio of $$\left|\frac{q_2}{q_3}\right|$$ is $$\frac{8}{5\sqrt{x}}$$, where $$x =$$ _____

At point P, net electric field along y-axis is zero.

So y-components of fields due to $$q_2\ and\ q_3$$​ must cancel.

Distance of P from charges:

For $$q_2$$​,

$$r_2=\sqrt{2^2+4^2}$$

$$=\sqrt{20}$$

For $$q_3$$​,

$$r_3=\sqrt{3^2+4^2}$$

=5

Electric field magnitude due to a point charge:

$$E=\frac{kq}{r^2}$$

Y-component is

$$E_y=E\sin⁡θ$$

For $$q_2$$​,

$$\sin\theta_2=\frac{4}{\sqrt{20}}$$

So

$$E_{2y}=\frac{kq_2}{20}\cdot\frac{4}{\sqrt{20}}$$

For $$q_3$$​,

$$\sin\theta_3=\frac{4}{5}$$

So

$$E_{3y}=\frac{kq_3}{25}\cdot\frac{4}{5}$$

For cancellation:

$$|E_{2y}|=|E_{3y}|$$

$$\frac{|q_2|}{20}\cdot\frac{4}{\sqrt{20}}=\frac{|q_3|}{25}\cdot\frac{4}{5}$$

Cancel 4:

$$\frac{|q_2|}{20\sqrt{20}}=\frac{|q_3|}{125}​$$

Thus

$$\frac{|q_2|}{|q_3|}=\frac{20\sqrt{20}}{125}$$

$$x=\frac{8}{5\sqrt{\ 5}}$$

so x=5