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We are given two particles fixed on an x-axis with charges and positions, and we are tasked with finding the point on the x-axis where the electric field produced by these particles is zero. Here's the step-by-step approach to solve this:

### Given:
- Charge \( q_1 = 2.07 \times 10^{-8} \, \text{C} \) at \( x_1 = 27.0 \, \text{cm} \).
- Charge \( q_2 = -4.41q_1 = -4.41 \times 2.07 \times 10^{-8} \, \text{C} \) at \( x_2 = 73.0 \, \text{cm} \).

### Step 1: Electric Field due to Point Charges
The electric field \( E \) at a point on the x-axis due to a point charge is given by:
\[
E = \frac{k \cdot |q|}{r^2}
\]
where:
- \( k \) is Coulomb's constant \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \),
- \( q \) is the charge,
- \( r \) is the distance from the charge to the point where we are calculating the electric field.

### Step 2: Net Electric Field
At some point \( x \) on the x-axis, the electric field due to both particles will be:
\[
E_1 = \frac{k \cdot |q_1|}{(x - x_1)^2}
\]
\[
E_2 = \frac{k \cdot |q_2|}{(x - x_2)^2}
\]
Since the electric field produced by \( q_1 \) and \( q_2 \) must cancel each other out, we set the magnitudes of the fields equal:
\[
\frac{|q_1|}{(x - x_1)^2} = \frac{|q_2|}{(x_2 - x)^2}
\]

### Step 3: Solve for \( x \)
Now substitute \( q_2 = -4.41q_1 \) and solve for \( x \):
\[
\frac{1}{(x - x_1)^2} = \frac{4.41}{(x_2 - x)^2}
\]

Substitute \( x_1 = 27.0 \, \text{cm} \) and \( x_2 = 73.0 \, \text{cm} \):
\[
\frac{1}{(x - 27)^2} = \frac{4.41}{(73 - x)^2}
\]

### Step 4: Simplification
Take the square root of both sides:
\[
\frac{1}{x - 27} = \frac{\sqrt{4.41}}{73 - x}
\]
\[
\frac{1}{x - 27} = \frac{2.1}{73 - x}
\]

Now, cross-multiply and solve for \( x \):
\[
73 - x = 2.1(x - 27)
\]
\[
73 - x = 2.1x - 56.7
\]
\[
73 + 56.7 = 2.1x + x
\]
\[
129.7 = 3.1x
\]
\[
x = \frac{129.7}{3.1} \approx 41.84 \, \text{cm}
\]

### Final Answer:
The electric field produced by the two charges is zero at approximately \( x = 41.84 \, \text{cm} \).

Would you like any further details or clarifications on this solution?

Here are some related questions:
1. What happens to the electric field between two charges of the same sign?
2. How would the solution change if both charges were positive?
3. Can we have multiple points where the electric field is zero with more charges?
4. What is the relationship between charge magnitude and electric field intensity?
5. How does the distance between two charges affect the zero-field point?

**Tip:** When calculating electric fields, always ensure to consider the vector nature, as the direction is crucial in determining where fields cancel out.

Two particles are fixed to an x-axis: particle 1 of charge q1 = 2.07 × 10^−8 C at x = 27.0 cm and particle 2 of charge q2 = −4.41q1 at x = 73.0 cm. At what coordinate on the x-axis is the electric field produced by the particles equal to zero?

Learn how to determine the point where the electric field produced by two charges is zero. This problem involves electrostatic concepts such as Coulomb's law and field cancellation, suitable for high school or introductory college-level physics.

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