B

Step 1: Understand the concept of electric potential (V). The electric potential due to a point charge is given by the formula:
V = k \frac{q}{r}
where:
- V is the electric potential,
- k is Coulomb's constant (8.99 \times 10^9 \text{ N m}^2/\text{C}^2),
- q is the charge,
- r is the distance from the charge.

Step 2: Consider the geometry of the problem. Let the position of A be at x = 0 and the position of B be at x = 0.6 m (60 cm). We want to find a distance x along the line AB where the potential is zero.

Step 3: The potential at point P (x) due to charges A and B can be expressed as:
V_P = V_A + V_B = k \frac{q_A}{r_A} + k \frac{q_B}{r_B}
where:
- q_A = 4 \times 10^{-6} C,
- q_B = -8 \times 10^{-6} C,
- r_A = x,
- r_B = 0.6 - x.

Step 4: Set the total potential to zero:
0 = k \frac{4 \times 10^{-6}}{x} + k \frac{-8 \times 10^{-6}}{0.6 - x}
canceling k, we get:
0 = \frac{4 \times 10^{-6}}{x} - \frac{8 \times 10^{-6}}{0.6 - x}

Step 5: Cross-multiply to solve for x:
4(0.6 - x) = 8x
2.4 - 4x = 8x
2.4 = 12x
x = \frac{2.4}{12} = 0.2 \, \text{m}

Step 6: Since the total potential can also be zero on the other side of the charges, consider points between them, and beyond the negative charge.
Step 7: Check location for potential zero on the left side of A or right side of B:
Left side (x < 0): V will not be zero, significant net positive.
Outside B (x > 0.6 m): Check V = k \frac{4 \times 10^{-6}}{x} + k \frac{-8 \times 10^{-6}}{x - 0.6} by similar procedure will yield potential zero at: x = 1.2 m

Conclusion: The points where the electric potential is zero are: 0.2 m and 1.2 m. Thus, the required points on the line AB are at these distances.