Question 21.67: In Fig. 21-26, particle 1 of charge -5.00q and particle 2 of......
In Fig. 21-26, particle 1 of charge -5.00q and particle 2 of charge +2.00q are held at separation L on an x axis. If particle 3 of unknown charge q_3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a) x and (b) y coordinates of particle 3?
THINK Our system consists of two charges along a straight line. We’d like to place a third charge so that the net force on it due to charges 1 and 2 vanishes.
EXPRESS The net force on particle 3 is the vector sum of the forces due to particles 1 and 2: \vec{F}_{3, \text { net }}=\vec{F}_{31}+\vec{F}_{32} . In order that \vec{F}_{3, \text { net }}=0 , particle 3 must be on the x axis and be attracted by one and repelled by another. As the result, it cannot be between particles 1 and 2, but instead either to the left of particle 1 or to the right of particle 2. Let q_3 be placed a distance x to the right of q_1 = -5.00q. Then its attraction to q _{1} particle will be exactly balanced by its repulsion from q_2 = +2.00q:
F_{3 x, \text { net }}=k\left[\frac{q_1 q_3}{x^2}+\frac{q_2 q_3}{(x-L)^2}\right]=k q_3 q\left[\frac{-5}{x^2}+\frac{2}{(x-L)^2}\right]=0 .
ANALYZE (a) Cross-multiplying and taking the square root, we obtain
\frac{x}{x-L}=\sqrt{\frac{5}{2}}
which can be rearranged to produce
x=\frac{L}{1-\sqrt{2 / 5}} \approx 2.72\, L .
(b) The y coordinate of particle 3 is y = 0.
LEARN We can use the result obtained above for consistency check. We find the force on particle 3 due to particle 1 to be
F_{31}=k \frac{q_1 q_3}{x^2}=k \frac{(-5.00 q)\left(q_3\right)}{(2.72 L)^2}=-0.675 \frac{k q q_3}{L^2} .
Similarly, the force on particle 3 due to particle 2 is
F_{32}=k \frac{q_2 q_3}{x^2}=k \frac{(+2.00 q)\left(q_3\right)}{(2.72 L-L)^2}=+0.675 \frac{k q q_3}{L^2} .
Indeed, the sum of the two forces is zero.