Answer:
FR₃=3385.3*10⁻⁹N(-j)
FR₃ in the direction of the y-axis and down
Explanation:
Theory of electrical forces
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)