Question

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 $x = \dots \dots \dots \dots \dots$.

Answer 1

The geometry of the charge configuration is shown in the figure, with $q_2$ located at a distance 2 cm and $q_3$ at 3 cm. The distances of these charges from point P are:

$\sqrt{20}$ cm and $\sqrt{25}$ cm.

The horizontal components of the electric field cancel each other, and the net electric field along the Y-axis is zero. Using the electric field formula:

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

the condition for $E_y = 0$ gives:

$\frac{kq_2}{20} \cos \beta = \frac{kq_3}{25} \cos \theta.$

The angles $\beta$ and $\theta$ are such that:

$\cos \beta = \frac{4}{\sqrt{20}}, \quad \cos \theta = \frac{4}{\sqrt{25}}.$

Substitute these into the equation:

$\frac{q_2}{20} \times \frac{4}{\sqrt{20}} = \frac{q_3}{25} \times \frac{4}{\sqrt{25}}.$

Simplify:

$\frac{q_2}{q_3} = \frac{20}{25} \times \frac{\sqrt{25}}{\sqrt{20}} = \frac{8}{5\sqrt{x}}.$

From the given condition:

$\sqrt{x} = \frac{8 \times 25 \times \sqrt{25}}{5 \times 20 \times \sqrt{20}}.$

Simplify:

$\sqrt{x} = \sqrt{5} \implies x = 5.$

Thus, the value of $x$ is:

$x = 5.$