Now let’s apply coulomb’s law and the superposition principle to calculate the force on a point charge due to the presence of other nearby charges. two point charges are located on the positive x axis of a coordinate system. charge q1=3.0nc is 2.0 cm from the origin, and charge q2 = -6.0 nc is 4.0 cm from the origin. what is the total force (magnitude and direction) exerted by these two charges on a third point charge q3=5.0nc located at the origin?

F = - 1.68 10⁻⁴ N

, it is directed to the left of the x-axis

Explanation:

Coulomb's law is

     F = k q₁ q₂ / r²

Where K is the Coulomb constant that value 8.99 10⁹ N m²/ C², q are the electric charges and r is the distance between them. Let's apply to our problem for each pair of charges

Let's reduce the magnitudes to the SI system

    q₁ = 3.0 nc (1C / 10 9 nC) = 3.0 10⁻⁹ C

    x₁ = 2.0 cm (1m / 100cm) = 2.0 10⁻² m

    q₂ = -6.0 nC = -6.0 10⁻⁹ C

    x₂ = 4.0 cm = 4.0 10⁻² m

    q₃ = 5.0 nC = 5.0 10⁻⁹ C

    x3 = 0 m

Charges q1 and q3

    r = x₁ -x₃

    r = 2.0 10⁻² -0

    r = 2.0 10⁻² m

    F₁₃ = 8.99 10⁹ 3.0 10⁻⁹ 5.0 10⁻⁹ / (2.0 10⁻²)²

    F₁₃ = 33.7 10⁻⁵ N

As the charges are of the same sign, the force is repulsive, therefore it is directed to the left of the x-axis

Charges q2 and q3

    r = r₂ –r₃

    r = 4.0 10⁻² - 0 = 4.0 10⁻² m

    F₂₃ = 8.99 10⁹ 6.0 10⁻⁹ 5.0 10⁻⁹ / (4.0 10⁻²)²

    F₂₃ = 16.86 10⁻⁵ N

As the charges are of different sign, the force is attractive, therefore it is directed to the right of the x-axis

The force is a vector magnitude, so each component must be added independently, in this case all the forces are on the x-axis, let's take the right direction as positive

    F = F₂₃ - F₁₃

    F = 16.86 10⁻⁵ - 33.7 10⁻⁵

    F = - 16.84 10⁻⁵ N

    F = - 1.68 10⁻⁴ N

The negative sign means that it is directed to the left of the x-axis