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The problem asks to determine the electric potential at 2 meters from a particle, given that the electric potential at 1 meter is 30 kV.

The formula for the electric potential \( V \) at a distance \( r \) from a point charge is:

\[
V(r) = \frac{k \cdot Q}{r}
\]

Where:
- \( V \) is the electric potential,
- \( k \) is Coulomb's constant,
- \( Q \) is the charge, and
- \( r \) is the distance from the charge.

From this, we can observe that the electric potential is inversely proportional to the distance from the charge. Therefore, if we double the distance, the potential will halve. Mathematically:

\[
V_2 = V_1 \cdot \frac{r_1}{r_2}
\]

Given:
- \( V_1 = 30 \, \text{kV} \),
- \( r_1 = 1 \, \text{m} \),
- \( r_2 = 2 \, \text{m} \),

We substitute into the formula:

\[
V_2 = 30 \, \text{kV} \cdot \frac{1 \, \text{m}}{2 \, \text{m}} = 15 \, \text{kV}
\]

Thus, the correct answer is **B) 15 kV**.

Would you like further details or have any questions?

Here are 5 related questions to explore:
1. What is the relationship between electric potential and distance from a point charge?
2. How is electric potential energy related to electric potential?
3. Can the electric potential ever be negative?
4. How does the presence of other charges affect electric potential?
5. What is Coulomb’s constant and how does it influence electric potential?

**Tip:** Remember, for point charges, the potential decreases as the distance increases due to the inverse proportionality.

El potencial eléctrico a 1 m de una partícula electrizada es 30 kV. Determine el potencial eléctrico a 2 m de la partícula.

Determine the electric potential at 2 meters from a charged particle when the potential at 1 meter is 30 kV. This physics problem involves concepts like Coulomb's Law and inverse proportionality. Suitable for Grades 10-12 students.

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