Question

A point charge q1 = -2.9 μC is located at the origin of a co-ordinate system. Another point charge q2 = 6.4 μC is located along the x-axis at a distance x2 = 7.6 cm from q1. Charge q2 is now displaced a distance y2 = 2.5 cm in the positive y-direction.A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 11.35 N and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3?How would you change q1 (keeping q2 and q3 fixed) in order to make the net force on q2 equal to zero?

Answer 1

Answer:

q3 = 1.04 μC    q1 = -4.16 μC

Explanation:

a) In order to find the value of the forces due to q1 and q3 on q2, we need to know first the distance between those charges and q2.

The coordinates for q2 are x₂ = 7.6 cm, y₂ = 2.5 cm.

The distance from the origin can be found applying the Pythagorean Theorem, as follows:

d₁₂ = √(7.6)²+(2.5)² = 8.0 cm

As q3 is positioned halfway between q1 (which is at the origin)  and q2 (which is at 8 cm. from the origin along a straight line), we conclude that q3 is located at 4 cm. from q2, and q1 at 8 cm. from q2 too.

According to Coulomb´s Law the force that both charges exert on q2, can be found applying superposition, as follows:

F₂ = F₁₂ + F₃₂ = 11.35 N

⇒ F₁₂ =9*10⁹ N*m²/C²*6.4*10⁻⁶μC*(-2.9 μC) /(0.0064)m² = -26.1 N

⇒F₃₂ = 11.35 N - (-26.1N) = 37.45 N

⇒ 37.45 N = 9*10⁹ N*m²/C²*6.4*10⁻⁶μC* q3 /(0.0016)m²

Solving for q3, we get:

q3 = (37.45 N / 36 N)* μC = 1.04 μC (q3 is a positive charge as expected).

b) Keeping q2 and q3 fixed, in order to make the net force on q2 equal to zero, we need to change the value of F₁₂ only, as follows:

F₁₂ + F₃₂ = 0 ⇒ F₁₂ = - F₃₂ = -37.45 N

⇒ -37.45 N = 9*10⁹ N*m²/C²*6.4*10⁻⁶μC*q1 /(0.0064)m²

⇒ q1 = (-37.45 N / 9 N) μC = -4.16 μC

As expected, we need that q1 be more negative in order to counteract the net force pointing away from q1.