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The problem in the image asks to find the position on the x-axis where the electric field produced by two charged particles is equal to zero. The information provided is:

- Particle 1: Charge \( q_1 = 2.98 \times 10^{-8} \, \text{C} \) at \( x = 25.0 \, \text{cm} \)
- Particle 2: Charge \( q_2 = -5.29q_1 \) at \( x = 67.0 \, \text{cm} \)

The question asks for the location on the x-axis where the electric field from these two particles cancels out, making the net electric field zero.

### Solution Outline:

1. **Electric Field Formula**: The electric field \( E \) created by a charge \( q \) at a distance \( r \) is given by:
   \[
   E = \frac{k \cdot |q|}{r^2}
   \]
   where \( k \) is Coulomb’s constant, \( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \), and \( r \) is the distance between the charge and the point of interest.

2. **Net Electric Field**: For the electric field to be zero, the magnitudes of the electric fields due to both charges must be equal but in opposite directions.

3. **Regions to Consider**: 
   - The point where the net electric field is zero could be located:
     - **To the left of particle 1** (i.e., \( x < 25.0 \, \text{cm} \))
     - **Between the two particles** (i.e., \( 25.0 \, \text{cm} < x < 67.0 \, \text{cm} \))
     - **To the right of particle 2** (i.e., \( x > 67.0 \, \text{cm} \))

4. **Set the Fields Equal**:
   Let \( x \) be the location where the net electric field is zero. The distances from \( x \) to the charges will differ depending on the region.

   - The distance from \( x \) to \( q_1 \) is \( |x - 25| \)
   - The distance from \( x \) to \( q_2 \) is \( |x - 67| \)

   For the electric fields to cancel out:
   \[
   \frac{k \cdot |q_1|}{(x - 25)^2} = \frac{k \cdot |q_2|}{(x - 67)^2}
   \]
   Simplifying this equation (since \( k \) cancels out):
   \[
   \frac{|q_1|}{(x - 25)^2} = \frac{|-5.29q_1|}{(x - 67)^2}
   \]
   \[
   \frac{1}{(x - 25)^2} = \frac{5.29}{(x - 67)^2}
   \]
   
5. **Solve the Equation**: Cross-multiply and solve for \( x \). After solving, the solution leads to:

   \[
   x = -7.3 \, \text{cm}
   \]

Thus, the electric field is zero at \( x = -7.3 \, \text{cm} \).

---

Do you want a more detailed breakdown of the algebra, or do you have any other questions? Here are some related questions for further understanding:

1. How does the sign of a charge affect the direction of the electric field?
2. What would happen if both charges had the same sign?
3. How do we know the field can't be zero between the charges?
4. What if the charges were placed on the y-axis instead of the x-axis?
5. How does the magnitude of charge affect the electric field at a distance?

**Tip:** In physics problems involving electric fields, always consider both magnitude and direction of the field contributions from different charges.

Two particles are fixed to an x-axis: particle 1 of charge q1 = 2.98 × 10^-8 C at x = 25.0 cm and particle 2 of charge q2 = -5.29q1 at x = 67.0 cm. At what coordinate on the x-axis is the electric field produced by the particles equal to zero?

Learn how to calculate the point on the x-axis where the electric field from two particles, one with charge q1 = 2.98 × 10^-8 C at x = 25.0 cm and another with charge q2 = -5.29q1 at x = 67.0 cm, cancels out. Detailed solution using Coulomb’s Law and algebraic techniques.

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