To find the point where the electric potential is zero, we use the formula for the electric potential due to a point charge:

To find the point where the electric potential is zero, we use the formula for the electric potential due to a point charge:

$$V = \frac{kq}{r}$$

where $k$ is Coulomb's constant, $q$ is the charge, and $r$ is the distance from the charge.

For two point charges, the total electric potential at a point is the algebraic sum of the potentials due to each charge:

$$V_{\text{total}} = V_1 + V_2 = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$$

We want to find the point where $V_{\text{total}} = 0$.

Given:
$q_1 = +4 \mu\text{C} = +4 \times 10^{-6} \text{ C}$
$q_2 = -2 \mu\text{C} = -2 \times 10^{-6} \text{ C}$
Distance between charges, $d = 30 \text{ cm}$

Let's consider a point on the line joining the charges. Since the charges have opposite signs, there are two regions where the potential can be zero:

1. Between the charges.
2. Outside the charges, on the side of the smaller magnitude charge (i.e., to the right of $q_2$).

Case 1: The point is between the charges.
Let the point be at a distance $x$ from $q_1$.
Then the distance from $q_2$ will be $r_2 = d - x = (30 - x) \text{ cm}$.
Setting the total potential to zero:

$$\frac{kq_1}{x} + \frac{kq_2}{d-x} = 0$$

Since $k$ is a non-zero constant, we can divide by $k$:

$$\frac{q_1}{x} + \frac{q_2}{d-x} = 0$$ $$\frac{q_1}{x} = -\frac{q_2}{d-x}$$

Substitute the values for $q_1$ and $q_2$:

$$\frac{+4 \times 10^{-6} \text{ C}}{x} = -\frac{-2 \times 10^{-6} \text{ C}}{30 \text{ cm} - x}$$ $$\frac{4}{x} = \frac{2}{30 - x}$$

Now, solve for $x$:

$$4(30 - x) = 2x$$ $$120 - 4x = 2x$$ $$120 = 6x$$ $$x = \frac{120}{6}$$ $$x = 20 \text{ cm}$$

This means the electric potential is zero at a point $20 \text{ cm}$ from the $+4 \mu\text{C}$ charge. This point is between the two charges.

Case 2: The point is outside the charges, to the right of $q_2$.
Let the point be at a distance $x$ from $q_1$.
Then the distance from $q_2$ will be $r_2 = x - d = (x - 30) \text{ cm}$.
Setting the total potential to zero:

$$\frac{kq_1}{x} + \frac{kq_2}{x-d} = 0$$ $$\frac{q_1}{x} = -\frac{q_2}{x-d}$$ $$\frac{+4 \times 10^{-6} \text{ C}}{x} = -\frac{-2 \times 10^{-6} \text{ C}}{x - 30 \text{ cm}}$$ $$\frac{4}{x} = \frac{2}{x - 30}$$ $$4(x - 30) = 2x$$ $$4x - 120 = 2x$$ $$2x = 120$$ $$x = 60 \text{ cm}$$

This means the electric potential is also zero at a point $60 \text{ cm}$ from the $+4 \mu\text{C}$ charge (which is $30 \text{ cm}$ to the right of the $-2 \mu\text{C}$ charge).

Comparing our results with the given options:
(A) 10 cm from +4µC
(B) 20 cm from +4µC
(C) 15 cm from +4µC
(D) 25 cm from [partially visible]

The value $x = 20 \text{ cm}$ from $+4 \mu\text{C}$ matches option (B).

The final answer is $\boxed{\text{B}}$.