Question

In Fig. 21-26, particles 1 and 2 are fixed in place on an xaxis, at a separation of$\mathit{L}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$.Their charges are ${\mathit{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathit{e}\mathbf{}$and${\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{27}\mathit{e}$. Particle 3 with charge${\mathit{q}}_{\mathbf{3}}\mathbf{=}\mathbf{+}\mathbf{4}$is to be placed on the line between particles 1 and 2, so that they produce a net electrostatic force${\overrightarrow{\mathbf{F}}}_{\mathbf{3}\mathbf{,}\mathbf{n}\mathbf{e}\mathbf{t}}$on it. (a) At what coordinate should particle 3 be placed to minimize the magnitude of that force? (b) What is that minimum magnitude?

Short answer

a)The coordinate value at which particle 3 needs to be placed to minimize the magnitude of that force is $2cm$.

b) The magnitude of the minimum force is $9.21x{10}^{-24}N$

Step-by-step solution

Stating the given data

1. Charges of the particles 1, 2 and 3 are$q_1=+e,q_2=-27eand{q}_3=+4e$.
2. The separation between particles 1 and 2 is $r=8cmor0.08cm$.

Understanding the concept of Coulomb’s law

Using the same concept of Coulomb's law, we can get the equation of the net force of the particles. Differentiating this equation will give the required value of x. Further, using this x-value, we can get the minimum force.

Formula:

The magnitude of the electrostatic force between any two particles is$F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$.… (i)

a) Calculation value of x that gives minimum force

Let $x$be the distance between particle 1 and particle 3. Thus, the distance between particle 3 and particle 2 is $(L-x)$. Both particles exert leftward forces on $q_3$(so long as it is on the line between them), so the magnitude of the net force using equation (i) on $q_3$is

$$\begin{gathered} F_{net}=\left|F_{13}\right|+\left|F_{23}\right| \\ =\frac{e^2}{\pi {\epsilon }_0}\left(\frac{1}{x^2}+\frac{27}{{\left(L-x\right)}^2}\right) \end{gathered}$$

Differentiating the above equation and equating it to zero, we can get the required valued of x as follows:

$$\begin{gathered} \frac{d}{dx}\left\{\frac{e^2}{\pi {\epsilon }_0}\left(\frac{1}{x^2}+\frac{27}{{\left(L-x\right)}^2}\right)\right\}=0 \\ \frac{-2x}{x^4}+\frac{2(L-x)27}{{(L-x)}^4}=0 \\ \frac{1}{x^3}=\frac{27}{{(L-x)}^3} \\ \frac{(L-x)}{x}=3 \\ \frac{L}{x}=4 \\ x=\frac{L}{4 \\ =\frac{8\text{cm}}{4} \\ =2\text{cm} \\ } \end{gathered}$$

Hence, the required value of x is $2cm$$2cm$.

b) Calculation of the minimum force

Substituting in the equation (a), we can get the net minimum force as

$$\begin{gathered} F_{net}=\frac{e^2}{\pi {\epsilon }_0}\left(\frac{1}{2^2}+\frac{27}{{\left(8-2\right)}^2}\right) \\ =9.21\times {10}^{-24}N \end{gathered}$$

Hence, the value of the minimum force is$9.21x{10}^{-24}N$