Two point charges, q_1 = 36 μC and q_2 = -9 μC are placed at a distan…

The correct answer is Option (3) → 60 cm

Two point charges $q_1 = 36\ \mu C$ and $q_2 = -9\ \mu C$ are separated by a distance $d = 0.3\ \text{m}$. The point where the net electric field is zero lies outside the line segment connecting the charges, on the side of the smaller magnitude charge ($q_2$).

Let the distance from $q_2$ to the zero-field point be $x$. Then the distance from $q_1$ to the zero-field point is $x + 0.3$.

The electric fields due to the charges must be equal in magnitude:

$\frac{|q_1|}{(x + 0.3)^2} = \frac{|q_2|}{x^2}$

Substitute values:

$\frac{36}{(x + 0.3)^2} = \frac{9}{x^2}$

Simplify:

$\frac{36}{9} = \frac{(x + 0.3)^2}{x^2}$

$4 = \frac{(x + 0.3)^2}{x^2}$

Take square root:

2 = $\frac{x + 0.3}{x}$

2x = x + 0.3

x = 0.3 m

Distance from $q_1$ to the zero-field point: x + 0.3 = 0.3 + 0.3 = 0.6 m

Final Answer: Distance from $q_1$ = 0.6 m