91Ó°ÊÓ

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 $r=8cmor0.08cm$.

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)

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

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

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

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

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

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

$$\begin{gathered} F_{net}=\frac{e^2}{\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{2^2}+\frac{27}{{\left(8-2\right)}^2}\right) \\ =9.21×{10}^{-24}N \end{gathered}$$

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