Acharge 4.96 nc is placed at the origin of an xy-coordinatesystem, and a charge -1.99 nc is placed on the positive x-axis at x = 4.01 cm . a third particle, of charge 5.99 nc is now placed at the point x = 4.01 cm , y = 2.98 cm . find the x-component of the total force exerted on the third charge by the other two. find the y-component of the total force exerted on the third charge by the other two. find the magnitude of the total force acting on the third charge. find the direction of the total force acting on the third charge.

B)  x-component of the total force exerted on the third charge by the other two (Fn₃x)

Fn₃x =8,56*10⁻⁵ N (+x)

B)  y-component of the total force exerted on the third charge by the other two (Fn₃y)

Fn₃y = 5.7*10⁻⁵ N  (-y)

Explanation:

Theory of electrical forces

Because the particle q₃ 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.

Equivalences:

1nC= 10⁻⁹C

1cm = 10⁻²m

Data

q₁=4.96 nC = +4.96*10⁻⁹C

q₂=-1.99 nC = -1.99*10⁻⁹C

d₂= 2.98 cm

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs.

 the force F₂₃ of q₂ and q₃ is attractive because the charges have opposite signs.

Calculation of the forces exerted on the charge q₁ and q₂ on q₃

To calculate the magnitudes of the forces exerted by the charges q₁, q₂, on q₃ we apply Coulomb's law:

F₁₃=k*q₁*q₃/d₁² =9*10⁹*4.96*10⁻⁹*5.99*10⁻⁹/(49.96*10⁻³)²=10.7**10⁻⁵ N

F₂₃=k*q₂*q₃/d₂² =9*10⁹*1.99*10⁻⁹*5.99*10⁻⁹/(2.98*10⁻²)²=12.08*10⁻⁵ N

F₁₃x=F₁₃cosβ=10.7*10⁻⁵* (4.01/4.996)=8,56*10⁻⁵ N  (+x)

F₁₃y=F₁₃sinβ= 10.7*10⁻⁵* (2.98/4.996)=6,38*10⁻⁵ N  (+x)

F₂₃x= 0

F₂₃y= F₂₃=12.08*10⁻⁵ N (-y)

A) x-component of the total force exerted on the third charge by the other two (Fn₃x)

Fn₃x= F₁₃x + F₂₃x= 8,56*10⁻⁵ N + 0

Fn₃x =8,56*10⁻⁵ N (+x)

B)  y-component of the total force exerted on the third charge by the other two (Fn₃y)

Fn₃y = F₁₃y + F₂₃y= 6,38*10⁻⁵ N - 12.08*10⁻⁵ N

Fn₃y = 5.7*10⁻⁵ N  (-y)

c. convergent boundary.

explanation:

for this one it would be d bump/forearm pass