Question

Two particles are fixed on an x-axis. Particle 1 of charge 40$\mu C$is located at x=-2.0cm; particle 2 of charge Qis located at x=3.0cm. Particle 3 of charge magnitude $20\mu C$is released from rest on the yaxis aty=2.0cm. What is the value of Qif the initial acceleration of particle 3 is in the positive direction of (a) the x-axis and (b) the y-axis?

Short answer

1. The value of Q on the x-axis in the initial acceleration of particle 3 is balanced .
2. The value of Q on the x-axis in the initial acceleration of particle 3 is $55.2\hspace{0.33em}\mu C$

Step-by-step solution

 Step 1: The given data

Particle 1 of charge$40\mu C$is located at x=-2.0cm

Particle 2 of charge Q is located at x=3.0cm

Particle 3 of charge 20$\mu C$is released from rest on the y axis at y=2.0cm

Understanding the concept of Coulomb’s law

Using the given formula of angle exerted by one force on the other force, we can get the angle between the particular forces. Hence, using this formula in the force equation of Coulomb's law, we can get the required components of acceleration by considering the required case of canceling the components to get the desired value.

Formulae:

The magnitude of the electrostatic force between any two particles,

$F_1=\frac{1}{\left(4\pi {\epsilon }_0\right)}\left(\frac{\left|q_1\right|\left|q_2\right|cos\theta }{r^2}i^+\frac{\left|q_1\right|\left|q_2\right|sin\theta }{r^2}j\right)$

(1)

The angle between two forces q₁,and q₂ $\theta ={\tan }^{-1}\left(\frac{r_1}{r_2}\right)$,$\theta ={\tan }^{-1}\left(\frac{r_1}{r_2}\right)$

a) Calculation of the value of Q in the positive x-axis

For the net force to be in the +x direction, the y components of the individual forces must cancel. The angle of the force exerted by the $q_1=40\hspace{0.33em}\mu C$charge on localid="1662639426410" $q_3=20\hspace{0.33em}\mu C$is localid="1663918785308" ${45}^{\circ }$, and the angle of force exerted on q₃ by Q is at –θ where the angle value is given using equation (2) as follows:

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{2.0\mathrm{cm}}{3.0\mathrm{cm}}\right)\\ =33.7^0\end{array}$

Therefore, the cancellation of y components is achieved if the net force of the y-component of particle 3 is balanced. This is found by substituting the given values in equation (1) of y-components to get the following equation:

localid="1662640231077" $k\frac{q_1{q}_3}{(0.02\surd 2m{)}^2}sin45°=k\frac{\left|Q\right|q_3}{\left(\sqrt{\left(\right(0.03m{)}^2+(0.02m{)}^2{)}_{}^2}\right)}sin33.7°$

|Q|=83$\mu C$

Charge Q is “pulling” on q₃ , so (since > 0) we conclude that is$-83\mu C$the charge value for the positive x-axis.

b) Calculation of value of Q in the positive y-axis

Now, we require that the x components cancel, and we note that in this case, the angle of force on exerted by Q is +θ (it is repulsive, and Q is positive-valued).

Therefore, the cancellation of x components is achieved if the net force of the y-component of particle 3 is balanced. This is found by substituting the given values in equation (1) of y-components to get the following equation:

$k\frac{q_1{q}_3}{(0.02\surd 2m{)}^2}\cos 45°=k\frac{\left|Q\right|q_3}{\left(\sqrt{\left(\right(0.03m{)}^2+(0.02m{)}^2{)}_{}^2}\right)}\cos 33.7°$

Q=55.2$\mu C$

Hence, the value of the charge in the positive y-axis is 55.2$\mu C$.