Let's break down the problem and solve for the net force acting on the third charge at the origin.

Given:

• Charge A: $3.0C$ at $(0,3)$
• Charge B: $9.0C$ at $(4,0)$
• Third charge (q) at the origin $(0,0)$

We'll assume that all charges are positive, meaning the third charge will experience repulsive forces from both Charge A and Charge B.

Step 1: Calculate Individual Forces

Using Coulomb's Law:

$F = k \frac{|q_1 q_2|}{r^2}$

where:

• $k = 9.0 \times 10^9$ $Nm^2/C^2$ (Coulomb's constant)
• $q_1$ and $q_2$ are the interacting charges
• $r$ is the distance between them

Force due to Charge A ($F_A$)

• Distance: $r_A = 3m$
• Charge: $3.0C$

$F_A = 9.0 \times 10^9 \times \frac{3.0 \times q}{3^2}$

$F_A = 9.0 \times 10^9 \times \frac{3.0 \times q}{9}$

$F_A = (9.0 \times 10^9) \times (0.33q) = 3.0 \times 10^9 q$

Direction: Since the force is repulsive, it acts downward (negative $y$-direction).

Force due to Charge B ($F_B$)

• Distance: $r_B = 4m$
• Charge: $9.0C$

$F_B = 9.0 \times 10^9 \times \frac{9.0 \times q}{4^2}$

$F_B = 9.0 \times 10^9 \times \frac{9.0 \times q}{16}$

$F_B = (9.0 \times 10^9) \times (0.5625q) = 5.06 \times 10^9 q$

Direction: The force is leftward (negative $x$-direction).

Step 2: Calculate Net Force

The forces are perpendicular to each other, so we use the Pythagorean theorem:

$F_{\text{net}} = \sqrt{F_A^2 + F_B^2}$

$F_{\text{net}} = \sqrt{(3.0 \times 10^9 q)^2 + (5.06 \times 10^9 q)^2}$

$F_{\text{net}} = \sqrt{9.0 \times 10^{18} q^2 + 25.6 \times 10^{18} q^2}$

$F_{\text{net}} = \sqrt{34.6 \times 10^{18} q^2}$

$F_{\text{net}} = 5.88 \times 10^9 q$

Step 3: Find Direction (Angle)

The angle $\theta$ with respect to the negative x-axis is:

$\theta = \tan^{-1} \left(\frac{F_A}{F_B}\right)$

$\theta = \tan^{-1} \left(\frac{3.0 \times 10^9}{5.06 \times 10^9}\right)$

$\theta = \tan^{-1} (0.593)$

$\theta \approx 30.6^\circ$

Final Answer:

• Magnitude of net force: $5.88 \times 10^9 q$ N
• Direction: $30.6^\circ$ below the negative x-axis (or in the third quadrant)

Would you like me to clarify anything further or provide additional details?

Related Questions:

1. What happens if one of the charges is negative?
2. How would the result change if the charges were in different locations?
3. How do electric field lines behave in this setup?
4. How do you calculate the force if the third charge is also moving?
5. How does this relate to vector addition in physics?

Tip:

Always break down vector forces into components before summing them up. This simplifies calculations significantly!