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

Steps to Solve

1. Identify the situation We have two positive charges, $q_1 = 15 , \mu C$ at $x = 2 , m$ and $q_2 = 6 , \mu C$ at the origin ($x = 0$). We need to find the position of the negative charge $q_3$ along the x-axis where the net electric force is zero.

2. Determine possible regions for $q_3$ The negative charge can be positioned in three possible regions:

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

3. Analyze forces on $q_3$ When placed at a point $x$, the forces acting on $q_3$ from $q_1$ and $q_2$ need to be equal in magnitude and opposite in direction for the net force to be zero.

4. Set up force equations Use Coulomb's Law, which states that the electric force $F$ between two charges is given by: $$ F = k \frac{|q_1 q_2|}{r^2} $$ where $k = 8.99 \times 10^9 , \text{N m}^2/\text{C}^2$.

For $q_3$ at position $x$, the magnitude of force due to $q_1$ is: $$ F_{q_1} = k \frac{|q_1 q_3|}{(x - 2)^2} $$

The magnitude of force due to $q_2$ is: $$ F_{q_2} = k \frac{|q_2 q_3|}{x^2} $$

Setting these equal gives: $$ k \frac{|q_1 q_3|}{(x - 2)^2} = k \frac{|q_2 q_3|}{x^2} $$

5. Simplify the equation Since $k$ and $|q_3|$ cancel out, we have: $$ \frac{q_1}{(x - 2)^2} = \frac{q_2}{x^2} $$

Substituting in the values for $q_1$ and $q_2$: $$ \frac{15}{(x - 2)^2} = \frac{6}{x^2} $$

6. Cross-multiply and solve for $x$ Cross-multiplying gives: $$ 15x^2 = 6(x - 2)^2 $$

Expanding the right side: $$ 15x^2 = 6(x^2 - 4x + 4) $$ $$ 15x^2 = 6x^2 - 24x + 24 $$

7. Rearranging and solving the quadratic equation Bringing all terms to one side: $$ 15x^2 - 6x^2 + 24x - 24 = 0 $$ $$ 9x^2 + 24x - 24 = 0 $$

Now we can apply the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a = 9$, $b = 24$, and $c = -24$.

8. Finding the roots Calculating the discriminant: $$ D = 24^2 - 4(9)(-24) = 576 + 864 = 1440 $$

Finding the roots: $$ x = \frac{-24 \pm \sqrt{1440}}{18} $$

9. Calculate the position(s) of $q_3$ Calculating $x$ gives the two possible values. Selecting the appropriate solution that falls within the relevant region.