Question

In Fig. 21-26, particle 1 of charge $\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathit{C}$and particle 2 of charge $\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathit{\mu }\mathit{C}\mathbf{}$are held at separation L=10.0cm on an x-axis. If particle 3 of unknown charge q₃is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a) xand (b) ycoordinates of particle 3?

Short answer

1. The x-coordinate of particle 3 for the net force on it to be zero is 14cm
2. The y-coordinate of particle 3 for the net force on it to be zero is 0cm

Step-by-step solution

The given data

The values of charges on the three particles are:

$q_1=+1.0\mu C,q_2=-3.0\mu C,\hspace{0.33em}and\hspace{0.33em}{q}_3=unknown$

The separation between particles 1 and 2,$L=10.0cm$

The net force on the particle from particles 1 and 2 is zero.

Understanding the concept of Coulomb’s law

Using the concept of Coulomb's law, we can get the required equation to solve to get the coordinates of particle 3 for the condition of the net force acting on it as zero.

Formula:

The magnitude of the electrostatic force between any two particles,

$F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$ (1)

a) Calculation of the x-coordinate of the location of particle 3

There is no equilibrium position for q₃ between the two fixed charges because it is being pulled by one and pushed by the other (since q₁and q₂ have different signs); in this region, this means the two force arrows on q₃are in the same direction and cannot cancel. It should also be clear that off-axis (with the axis defined as that which passes through the two fixed charges) there are no equilibrium positions. On the semi-infinite region of the axis that is nearest q₂and furthest from q₁. an equilibrium position for q₃cannot be found because |q₁|$<$|q₂| and the magnitude of the force exerted by q₂is everywhere (in that region) stronger than that exerted by q₁on q₃. Thus, we must look in the semi-infinite region of the axis which is nearest q₁and furthest from q₂, where the net force on q₃has magnitude using equation (1) as follows:

$F_3=F_{13}+F_{23}$

$=\left|k\frac{\left|q_1{q}_3\right|}{L_0^2}-k\frac{\left|q_2{q}_3\right|}{{(L+L_0)}^2}\right|.....................\left(2\right)$

with L = 10 cm and L₀ is assumed to be positive. We set this equal to zero, as required by the problem. Now, we obtain the equation (2) by canceling the equal terms as given:
$$\begin{gathered} \frac{\left|q_1\right|}{\left(L_0^2\right)}-\frac{\left|q_2\right|}{(L+L_0{)}^2}=0 \\ {\left(\frac{L+L_0}{L_0}\right)}^2=\left|\frac{q_2}{q_1}\right| \\ {\left(\frac{L+L_0}{L0}\right)}^2=\left|\right(-3.0\mu C)/(-1.0\mu C\left)\right| \\ {\left(\frac{L+L_0}{L_0}\right)}^{}=3.0 \end{gathered}$$

Therefore, we get,

localid="1662643008359" $$\begin{gathered} (L+L_0)/L_0=\surd 3 \\ {L}_0=\frac{L}{(\surd 3-1)} \\ {L}_0=\frac{10cm}{(\surd 3-1)} \end{gathered}$$

L₀=14cm

Hence, the value of the x-coordinate of particle 3 is 14cm

b) Calculation of the y-coordinate of the location of particle 3

As stated above in the calculations of part (a), y = 0

Hence, the value of the y-coordinate of particle 3 is 0cm.