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

Correct Answer: 5

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:

\[ x = 5. \]