Question

In Fig. 21-25, particle 1 of charge &1.0 mC and particle 2 of charge $3.0 mC are held at separation L ! 10.0 cm on an x axis. If particle 3 of unknown charge q3 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?

Answer 1

Answer:

Since the particle 1 and 2 are on the x-axis, the 3rd particle should also be on the x-axis in order the net force on it to be zero.

Let's denote the distance between particles 1 and 3 as x. Therefore the distance between particles 2 and 3 is (0.1 - x), since the distance between 1 and 2 is 0.1 m.

Coulomb's Law states the force between charges as

$F_{1-3} = \frac{1}{4\pi \epsilon_0}\frac{q_1q_3}{x^2}$

$F_{2-3} = \frac{1}{4\pi \epsilon_0}\frac{q_2 q_3}{(0.1 -x)^2}$

The question asks that $F_{1-3} = F_{2-3}$, so

$\frac{1}{4\pi \epsilon_0}\frac{1\times 10^{-3} \times q_3}{x^2} = \frac{1}{4\pi \epsilon_0}\frac{3\times 10^{-3}\times q_3}{(0.1 - x)^2}\\\frac{1}{x^2} = \frac{3}{(0.1 - x)^2}\\3x^2 = 0.01 -0.2x + x^2\\2x^2 + 0.2x - 0.01 = 0\\x_1 = 0.036m\\x_2 = -0.136m$

We will take the positive root:

$x = 0.036~m$ away from the first particle.

Explanation:

Since Fig. 21-25 is not given in the question, the exact locations are not known. However, the location of the third particle is found to be 0.036 m away from the first particle and the third particle is located between the particles 1 and 2.