Question

a charge $q_1 = 5.00$ nc is placed at the origin of an xy - coordinate system, and a charge $q_2 = - 2.50$ nc is placed on the positive x axis at $x = 4.00$ cm.
part a
if a third charge $q_3 = 2.50$ nc is now placed at the point $x = 4.00$ cm, $y = 3.00$ cm, find the x component of the total force exerted on this charge by the other two charges.
express your answer in newtons.

Explanation:

Step1: Convert units to SI

Convert all distances to meters and charges to coulombs:
$q_1 = 5.00\ \text{nC} = 5.00 \times 10^{-9}\ \text{C}$
$q_2 = -2.50\ \text{nC} = -2.50 \times 10^{-9}\ \text{C}$
$q_3 = 2.50\ \text{nC} = 2.50 \times 10^{-9}\ \text{C}$
$r_{13} = \sqrt{(0.04\ \text{m})^2 + (0.03\ \text{m})^2} = 0.05\ \text{m}$
$r_{23} = 0.03\ \text{m}$
Coulomb's constant $k = 8.988 \times 10^9\ \text{N·m}^2/\text{C}^2$

Step2: Force from $q_1$ on $q_3$

Calculate magnitude and x-component:
Magnitude: $F_{13} = k\frac{|q_1 q_3|}{r_{13}^2} = 8.988 \times 10^9 \times \frac{5.00 \times 10^{-9} \times 2.50 \times 10^{-9}}{(0.05)^2}$
$F_{13} = 4.494 \times 10^{-5}\ \text{N}$
x-component: $F_{13x} = F_{13} \times \frac{0.04}{0.05} = 4.494 \times 10^{-5} \times 0.8 = 3.5952 \times 10^{-5}\ \text{N}$

Step3: Force from $q_2$ on $q_3$

Calculate magnitude and x-component (force is along +x, since $q_2$ is negative, $q_3$ is positive):
Magnitude: $F_{23} = k\frac{|q_2 q_3|}{r_{23}^2} = 8.988 \times 10^9 \times \frac{2.50 \times 10^{-9} \times 2.50 \times 10^{-9}}{(0.03)^2}$
$F_{23} = 6.243 \times 10^{-5}\ \text{N}$
x-component: $F_{23x} = 6.243 \times 10^{-5}\ \text{N}$

Step4: Sum x-components

$F_x = F_{13x} + F_{23x} = 3.5952 \times 10^{-5} + 6.243 \times 10^{-5}$

Answer:

$9.84 \times 10^{-5}\ \text{N}$