Question

Two charged particles, with charges q1=qq1=q and q2=4qq2=4q, are located on the x axis separated by a distance of 2.00cm2.00cm . A third charged particle, with charge q3=qq3=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3.Find the position of charge 3 when qqq = 2.00 nCnC .

Answer 1

Answer: Charge 3 is located on the x-axis a distance of 0.67 cm from charge 1 and 1.33 cm from charge 2.

Explanation: Electrostatic Force is the force of repulsion or attraction between two charged particles. It's directly proportional to the charge of the particles and inversely proportional to the distance between them:

$F=k\frac{|q||Q|}{r^{2}}$

k is an electrostatic constant

For the system of 3 particles, suppose distance from 1 to 3 is x meters, so, distance from 2 to 3 is (0.02-x) meters.

Force will be

$F_{13}=F_{23}$

$k\frac{q_{1}q_{3}}{r_{13}^{2}} =k\frac{q_{2}q_{3}}{r_{23}^{2}}$

$\frac{q_{1}}{r_{13}^{2}} =\frac{q_{2}}{r^{2}_{{23}}}$

Substituting:

$\frac{2.10^{-6}}{x^{2}} =\frac{8.10^{-6}}{(0.02-x)^{2}}$

$8.10^{-6}x^{2}=2.10^{-6}(0.0004-0.04x+x^{2})$

$4x^{2}=x^{2}-0.04x+0.0004$

$3x^{2}+0.04x-0.0004=0$

Solving quadratic equation using Bhaskara:

$x_{1}=\frac{-0.04+\sqrt{(0.04)^{2}+0.048} }{6}$

$x_{2}=\frac{-0.04-\sqrt{(0.04)^{2}+0.048} }{6}$

x₂ will give a negative value and since distance can't be negative, use x₁

$x_{1}=\frac{-0.04+\sqrt{0.0064} }{6}$

x₁ = 0.0067 m

The position of charge 3 is 0.67 cm from charge 1 and 1.33 cm from charge 2.