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We are tasked with finding the point along the x-axis where the electric field due to two particles becomes zero. Here's how we can approach the problem step by step.

### Given Data:
- **Particle 1:** Charge \( q_1 = 2.1 \times 10^{-8} \, C \) located at \( x = 20 \, cm \)
- **Particle 2:** Charge \( q_2 = -4q_1 = -4 \times 2.1 \times 10^{-8} = -8.4 \times 10^{-8} \, C \) located at \( x = 70 \, cm \)
- We need to find a point along the x-axis where the electric field produced by these charges cancels out.

### Electric Field Formula:
The electric field \( E \) produced by a charge \( q \) at a distance \( r \) is given by:
\[
E = k \frac{q}{r^2}
\]
where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant.

### Step 1: Setup for Zero Electric Field
Let the point where the electric field is zero be at a distance \( x \) from the origin. There are three possible regions where the electric field could be zero:
1. **To the left of particle 1** (\( x < 20 \, cm \))
2. **Between particle 1 and particle 2** (\( 20 \, cm < x < 70 \, cm \))
3. **To the right of particle 2** (\( x > 70 \, cm \))

We will analyze the second region (between 20 cm and 70 cm) because it is most likely the field cancels out between the two charges since they have opposite signs.

Let \( d_1 = x - 20 \, cm \) and \( d_2 = 70 \, cm - x \) be the distances from the point where the field is zero to particles 1 and 2, respectively.

### Step 2: Set the Electric Fields Equal in Magnitude
For the electric field to be zero, the magnitude of the electric fields due to the two charges must be equal, i.e.,
\[
E_1 = E_2
\]
Substituting the electric field formula:
\[
k \frac{q_1}{d_1^2} = k \frac{q_2}{d_2^2}
\]
The \( k \)'s cancel out, and substituting the values of \( q_1 \) and \( q_2 \):
\[
\frac{q_1}{(x - 20)^2} = \frac{-4q_1}{(70 - x)^2}
\]
Simplifying:
\[
\frac{1}{(x - 20)^2} = \frac{4}{(70 - x)^2}
\]
Taking the square root of both sides:
\[
\frac{1}{x - 20} = \frac{2}{70 - x}
\]
Cross-multiplying:
\[
70 - x = 2(x - 20)
\]
Expanding:
\[
70 - x = 2x - 40
\]
Bringing like terms together:
\[
70 + 40 = 2x + x
\]
\[
110 = 3x
\]
\[
x = \frac{110}{3} \approx 36.67 \, cm
\]

Thus, the electric field is zero at approximately \( x = 36.67 \, cm \).

Would you like further details or clarification on any part of this solution?

### Follow-up Questions:
1. How would the result change if both charges were positive?
2. What if the charges were of equal magnitude but opposite in sign?
3. Could there be multiple points where the electric field is zero in some cases?
4. How does the electric field behave infinitely far away from both charges?
5. Can you derive the electric potential along the x-axis for this system?

#### Tip: 
The point where the electric field is zero typically lies closer to the smaller charge.

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.1 × 10−8 C at x = 20 cm and particle 2 of charge q2 = -4.00q1 at x = 70 cm. At what coordinate on the x axis is the electric field produced by the particles equal to zero?

Find the coordinate on the x-axis where the electric field between two charges is zero. Given q1 = 2.1 × 10−8 C and q2 = -4.00q1, this problem uses Coulomb's law and algebraic manipulation to solve for the point of zero electric field. Ideal for undergraduate physics students.

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