Two point charges are located on the y-axis as follows: one charge q1 = -1.40 nC located at y= -0.555 m, and a second charge q2 = 3.30 nC at the origin (y=0). What is the magnitude of the total force exerted by these two charges on a third charge q3 = 5.15nC located at y3 = -0.390m ?

Because the particle 3 is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Graphic attached

The directions of the individual forces exerted by q1 and q2 on q3 are shown in the attached figure; the force (F23) of q2 on q3 is repulsive because they have equal signs and the force (F13) of q1 on q3 is attractive because they have opposite signs.

Forces F13 and F23 are directed down on the y axis.

To calculate the magnitudes of the forces we apply Coulomb's law:

F₁₃=k*q₁*q₃/(r₁₃)² Equation (1): Magnitude of the electrical force of q₁ over q₃.

F₂₃=k*q₂*q₃/(r₂₃)² Equation (2) : Magnitude of the electrical force of q₂ over q₃.

Equivalences

1nC= 10-⁹ C

Known data

q₁=-1.4 nC=-2.5 *10-⁹C

q₂=3.3 nC=2.5 *10-⁹C

q₃=5.15 nC=2.5*10-⁹C

r₂₃ =0.390m: distancia entre q₂ y q₃

r₁₃= 0.555-0.390=0.165: distancia entre q₁ y q₃

k=8.99x10⁹N*m²/C² : Coulomb constant

F₁₃ calculation

We replace data in the equation( 1):

F₁₃=8,99*10⁹*1.4*10⁻⁹*5.15*10⁻⁹/0.165²

F₁₃=2380.8*10⁻⁹N, in the direction of the y-axis and down

F₂₃ calculation

We replace data in the equation (2):

F₂₃=8,99*10⁹*3.3*10⁻⁹*5.15*10⁻⁹/0.390²

F₂₃=1004.5*10⁻⁹N , in the direction of the y-axis and down

Calculation of the resulting force on q₃: FR3₃

We do the algebraic sum of F₁₃ and F₂₃ because both forces have the same direction.

FR₃=F₁₃+F₂₃=2380.8*10⁻⁹N +1004.5*10⁻⁹N =3385.3*10⁻⁹N: in the direction of the y-axis and down

FR₃=3385.3*10⁻⁹N(-j)