Question

Find the net force on a \(+2.00-\mathrm{C}\) charge at the origin of an \(x y\) -coordinate system if there is \(a+5.00-C\) charge at \((3.00 \mathrm{~m}, 0.00)\) and a \(-3.00-\mathrm{C}\) charge at \((0.00 .4 .00 \mathrm{~m})\)

Short answer

Solution: After applying Coulomb's Law and calculating the individual forces, the net force acting on the \(+2.00C\) charge at the origin is found by using the Pythagorean theorem: $$F_{net} = \sqrt{F_{net_x}^2 + F_{net_y}^2}$$

Step-by-step solution

Determine the distance between the charges

First, we need to find the distance between the \(+2.00C\) charge at the origin and the other two charges. The distance between the \(+2.00C\) charge and the \(+5.00C\) charge is 3 meters along the x-axis, while the distance between the \(+2.00C\) charge and the \(-3.00C\) charge is 4 meters along the y-axis.

Calculate the force between each pair of charges

To find the force between each pair of charges, we will use Coulomb's Law: $$ F = k\frac{q_1q_2}{r^2} $$ where F is the force between two charges, \(q_1\) and \(q_2\) are the values of the charges, r is the distance between the charges, and k is the electrostatic constant, which is approximately \(8.99 \times 10^9 Nm^2C^{-2}\). We will first calculate the force between the \(+2.00C\) and \(+5.00C\) charges: $$F_1 = \frac{(8.99 \times 10^9)(2)(5)}{3^2}$$ and then the force between the \(+2.00C\) and \(-3.00C\) charges: $$F_2 = \frac{(8.99 \times 10^9)(2)(-3)}{4^2}$$

Determine the direction of the individual forces

Since the \(+5.00C\) charge is at \((3.00m, 0.00)\), the force between this charge and the \(+2.00C\) charge at the origin will be in the negative x-direction. The \(-3.00C\) charge is at \((0.00, 4.00m)\), so the force between this charge and the \(+2.00C\) charge at the origin will be in the positive y-direction.

Calculate the x and y components of the force

Now we need to determine the x and y components of the force. Since the force from the \(+5.00C\) charge is entirely in the x-direction, we don't need to break it up into components. However, we do need to find the y-component of the force from the \(-3.00C\) charge. Since the force is in the positive y-direction and there is no x-component, we can just use \(F_2\) as the y-component of the force.

Add the x and y components of the forces

To find the net force on the \(+2.00C\) charge, we simply add the x-components and y-components of the individual forces. For the x-component, we have the force from the \(+5.00C\) charge: $$F_{net_x} = F_1$$ For the y-component, we have the force from the \(-3.00C\) charge: $$F_{net_y} = F_2$$

Find the net force

Finally, we can use the Pythagorean theorem to find the magnitude of the net force: $$F_{net} = \sqrt{F_{net_x}^2 + F_{net_y}^2}$$ After calculating the magnitudes of the individual forces and the net force, we can find the net force acting on the \(+2.00C\) charge at the origin due to the other two charges.