Answer to Question #133236 in Mechanics | Relativity for Jessica

Question

Two point charges are fixed on the y axis: a negative point charge q1 = -27 μC at y1 = +0.21 m and a positive point charge q2 at y2 = +0.37 m. A third point charge q = +8.8 μC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 25 N and points in the +y direction. Determine the magnitude of q2.

Answer 1

Explanations & Calculations

• Force between q₁ & q is attractive since those are opposite charges.
• That force can be calculated to be,

$\qquad\qquad \begin{aligned} \small F_{q_1q} &= \small \frac{1}{4 \pi \epsilon_0} \times\frac{|q_1||q|}{(0.21m)^2}\\ &= \small 8.9875\times 10^{-3}\times \frac{27\times 8.8}{0.0441}\\ &= \small 48.4224 \,\,N \end{aligned}$

• Force between q₂ & q is repulsive because of the like charges & that is,

$\qquad\qquad \begin{aligned} \small F_{q_2q} &= \small 8.9875\times 10^{-3}\times \frac{8.8\times q_2}{0.37^2}\\ &= \small 0.5777q_2 \,\,N \end{aligned}$

• Since the net force is directed along +y, $\small F_{q_1q}> F_{q_2q}$ therefore,

$\qquad\qquad \begin{aligned} \small F_{q_1q}-F_{q_2q} &= \small 25 \,\,N\\ \small 48.4224-0.5777q_2 &= \small 25\\ \small q_2 &= \small \bold{40.5442}=\bold{40.5\mu C} \end{aligned}$