Respuesta :
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
[tex]F_{1-3} = \frac{1}{4\pi \epsilon_0}\frac{q_1q_3}{x^2}[/tex]
[tex]F_{2-3} = \frac{1}{4\pi \epsilon_0}\frac{q_2 q_3}{(0.1 -x)^2}[/tex]
The question asks that [tex]F_{1-3} = F_{2-3}[/tex], so
[tex]\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[/tex]
We will take the positive root:
[tex]x = 0.036~m[/tex] 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.
