Two point charges $q_1 = 10 × 10^{-8} C$ and $q_2 = -2× 10^{-8} C$ are separated by a distance of 6 cm. The distance from $q_1$ where the electric potential is zero, can be

5 cm and 7.5 cm

The correct answer is Option (2) → 5 cm and 7.5 cm

Charges: $q_1 = +10 \times 10^{-8} \, C$, $q_2 = -2 \times 10^{-8} \, C$

Separation = $6 \, cm$

Case 1: Point between the charges (distance $x$ from $q_1$)

$\frac{10}{x} = \frac{2}{6 - x}$

$10(6 - x) = 2x$

$60 - 10x = 2x$

$60 = 12x$

$x = 5 \, cm$

Case 2: Point outside, on the side of $q_1$ (distance $x$ from $q_1$, other charge at $x+6$)

$\frac{10}{x} = \frac{2}{x + 6}$

$10(x + 6) = 2x$

$10x + 60 = 2x$

$8x = -60 \;\; \Rightarrow \;\; x = -7.5 \, cm$ (not possible since distance cannot be negative)

Case 3: Point outside, on the side of $q_2$ (distance $x$ from $q_1$, so from $q_2$ it is $x - 6$)

$\frac{10}{x} = \frac{2}{x - 6}$

$10(x - 6) = 2x$

$10x - 60 = 2x$

$8x = 60$

$x = 7.5 \, cm$ (from $q_1$, beyond $q_2$)

Final Answer: Electric potential is zero at $5 \, cm$ (between charges) and at $7.5 \, cm$ (outside, beyond $q_2$).