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The problem provides the electric potential \( V(x, y) \) at any point \((x, y)\), given by the expression:

\[
V(x, y) = \frac{5}{\sqrt{25 + x^2 + y^2}}
\]

We are asked to draw the equipotential curves for the values \( V = \frac{1}{2}, V = \frac{1}{3}, \text{ and } V = \frac{1}{4} \).

### Step-by-step approach:

1. **Equipotential Curves**:
   The equipotential curves correspond to points where \( V(x, y) \) is constant. To find these curves, we set the given potential values equal to the expression for \( V(x, y) \).

2. **Case 1: \( V = \frac{1}{2} \)**
   
   Setting \( V = \frac{1}{2} \), we solve for \( x^2 + y^2 \):
   \[
   \frac{1}{2} = \frac{5}{\sqrt{25 + x^2 + y^2}}
   \]
   Squaring both sides:
   \[
   \left( \frac{1}{2} \right)^2 = \frac{25}{25 + x^2 + y^2}
   \]
   \[
   \frac{1}{4} = \frac{25}{25 + x^2 + y^2}
   \]
   Cross-multiplying:
   \[
   25 + x^2 + y^2 = 100
   \]
   \[
   x^2 + y^2 = 75
   \]
   This is a circle with radius \( \sqrt{75} = 5\sqrt{3} \) centered at the origin.

3. **Case 2: \( V = \frac{1}{3} \)**
   
   Similarly, for \( V = \frac{1}{3} \):
   \[
   \frac{1}{3} = \frac{5}{\sqrt{25 + x^2 + y^2}}
   \]
   Squaring both sides:
   \[
   \left( \frac{1}{3} \right)^2 = \frac{25}{25 + x^2 + y^2}
   \]
   \[
   \frac{1}{9} = \frac{25}{25 + x^2 + y^2}
   \]
   Cross-multiplying:
   \[
   25 + x^2 + y^2 = 225
   \]
   \[
   x^2 + y^2 = 200
   \]
   This is a circle with radius \( \sqrt{200} = 10\sqrt{2} \).

4. **Case 3: \( V = \frac{1}{4} \)**
   
   For \( V = \frac{1}{4} \):
   \[
   \frac{1}{4} = \frac{5}{\sqrt{25 + x^2 + y^2}}
   \]
   Squaring both sides:
   \[
   \left( \frac{1}{4} \right)^2 = \frac{25}{25 + x^2 + y^2}
   \]
   \[
   \frac{1}{16} = \frac{25}{25 + x^2 + y^2}
   \]
   Cross-multiplying:
   \[
   25 + x^2 + y^2 = 400
   \]
   \[
   x^2 + y^2 = 375
   \]
   This is a circle with radius \( \sqrt{375} = 5\sqrt{15} \).

### Conclusion:

The equipotential curves for the given potential values are circles centered at the origin with the following radii:
- For \( V = \frac{1}{2} \), the radius is \( 5\sqrt{3} \).
- For \( V = \frac{1}{3} \), the radius is \( 10\sqrt{2} \).
- For \( V = \frac{1}{4} \), the radius is \( 5\sqrt{15} \).

Let me know if you would like a more detailed explanation, further clarifications, or help with drawing these curves. Here are some follow-up questions for better understanding:

1. What is the physical significance of an equipotential curve?
2. How does the radius of the equipotential curve change with the value of the potential \( V \)?
3. Can equipotential curves intersect each other? Why or why not?
4. What happens to the potential \( V \) as \( x \) and \( y \) increase infinitely?
5. How can the electric field be derived from the potential function \( V(x, y) \)?

**Tip**: The electric field at any point is always perpendicular to the equipotential surfaces at that point.

El potencial eléctrico en cualquier punto (x, y) es V(x, y) = 5 / sqrt(25 + x^2 + y^2). Dibujar las curvas equipotenciales de V = 1/2, V = 1/3 y V = 1/4.

Learn how to draw equipotential curves for the electric potential V = 5 / sqrt(25 + x^2 + y^2) at various potential values. This problem explores the concept of electric potential and equipotential surfaces, relevant to university-level physics.

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