Where must a negative charge (q3) be placed on the x-axis so that the resultant electric force on it is zero, considering q1 = 15 µC at x = 2 m and q2 = 6 µC at the origin?

Steps to Solve

1. Identify the positions of the charges The positive charges are positioned as follows:

• Charge $q_2 = 6 , \mu C$ is at $x = 0 , m$ (the origin)
• Charge $q_1 = 15 , \mu C$ is at $x = 2 , m$

2. Determine the location of charge $q_3$ Let the position of the negative charge $q_3$ be at $x = d$ along the x-axis. We will consider two possible regions for $d$:

• To the left of $q_2$ (i.e., $d < 0$)
• Between $q_2$ and $q_1$ (i.e., $0 < d < 2$)
• To the right of $q_1$ (i.e., $d > 2$)

3. Analyze the forces from the charges For the charge $q_3$ to experience zero net electric force, the magnitudes of the forces exerted by both charges must be equal.

The electric force $F$ exerted by a point charge is given by: $$ F = k \frac{|q_1 q_2|}{r^2} $$ Where $k = 8.99 \times 10^9 , \text{N m}^2/\text{C}^2$ is the Coulomb's constant and $r$ is the distance between the charges.

4. Calculate distances for both cases

• If $d < 0$:

  • Distance from $q_2$ to $q_3$ is $|d|$
  • Distance from $q_1$ to $q_3$ is $|d - 2| = 2 + |d|$ (since $d$ is negative)

• If $d > 2$:

  • Distance from $q_1$ to $q_3$ is $d - 2$
  • Distance from $q_2$ to $q_3$ is $d$

5. Set up equations for each case For $d < 0$: $$ |F_2| = |F_1| $$ $$ k \frac{6 \times 10^{-6} |q_3|}{d^2} = k \frac{15 \times 10^{-6} |q_3|}{(2 + |d|)^2} $$

For $d > 2$: $$ |F_2| = |F_1| $$ $$ k \frac{6 \times 10^{-6} |q_3|}{d^2} = k \frac{15 \times 10^{-6} |q_3|}{(d-2)^2} $$

6. Solve the equations for both cases Since $k$ and $|q_3|$ cancel out: For $d < 0$: $$ 6(2 + |d|)^2 = 15d^2 $$

For $d > 2$: $$ 6d^2 = 15(d - 2)^2 $$

7. Find potential solutions

• Simplify and solve the equations derived above.