Question

The charges and coordinates of two charged particles held fixed in an xy plane are q₁ = +2.4 μC, x₁ = 3.5 cm, y₁ = 0.50 cm and q₂ = −4.2 μC, x₂ = −2.0 cm, y₂ = 1.5 cm.

(a) Find the magnitude and direction of the electrostatic force on particle 2 due to particle 1.

(b) At what x and y coordinates should a third particle of charge q₃ = +3.9 μC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?

Answer 1

a)

Distance Between q₂ and q₁ is

$r=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}=\sqrt{(-2-3.5)^{2}+(1.5-0.5)^{2}}=5.6 cm$

Magnitude of Electrostaic force on particel 2 due to particle 1 is

$F=\frac{K|q_{1}||q_{2}|}{r^{2}}=\frac{(9\times 10^{9})(2.4\times 10^{-6})(-4.2\times 10^{-6})}{0.056^{2}}=29.03 N$

Since q₂ is attracted to q₁ ,the direction of the force is same as the vector points from q₂ to q₁ .so

So direction of force is

$\theta =tan^{-1}\left ( \frac{-1}{5.5} \right )=-10.3^{o}$

b)

The Force exerted by charge q₃=3.9 uC on q₂ is equal to Force exerted on q₂ by q₁

$F_{12}=F_{32}$

$29.03=\frac{K|q_{2|}|q_{3}|}{r_{o}^{2}}$

$29.03=\frac{(9\times 10^{9})(4.2\times 10^{-6})(3.9\times 10^{-6})}{r_{o}^{2}}$

So direction is

$\theta _{o}=180-10.3 =169.7^{o}$

The Charge q₃ should be located at

$x_{3}=x_{2}+r_{o}Cos\theta _{o}=-2+7.13cos169.7=-9.01 cm$

$y_{3}=y_{2}+r_{o}Sin\theta _{o}=1.5+7.13Sin169.7=2.77 cm$