91Ó°ÊÓ

A charge of $\mathrm{Q}=6.0\text{μ°ä´Ç°ù}6×{10}^{−6}\text{C}$ is to be split into two parts that are then separated by$\mathrm{r}=3.0\text{mmor}0.003\text{″¾}$

Using the concept of Coulomb's law, we can find the magnitude of the required force between the particles. Now, equating the derivation of the force with respect to charge as zero can give us the maximum possible value of charge to be split to give the maximum force.

Formula:

The magnitude of the electrostatic force between any two particles,

$\mathrm{F}=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$ (i)

Let, the charge splits up into charges of values,$q_1$and$\mathrm{Q}−{\mathrm{q}}_1$

The, using equation (i), the net electrostatic force acting between the parts can be given as:

$\mathrm{F}=\frac{{\mathrm{q}}_1(\mathrm{Q}−{\mathrm{q}}_1)}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\mathrm{................................}\left(\mathrm{a}\right)$

To get the maximum possible values of charges, we take the derivative of equation (i) with respect to charge ${\mathrm{q}}_1$,and equating it to zero, we get the charge value as:

$\begin{array}{c}\frac{\mathrm{d}}{{\mathrm{dq}}_1}\left(\frac{{\mathrm{q}}_1(\mathrm{Q}−{\mathrm{q}}_1)}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\right)=0\\ \mathrm{Q}−2{\mathrm{q}}_1=0\\ {\mathrm{q}}_1=\mathrm{Q}/2\end{array}$

Substituting this value of charge in equation (a), we get the maximum force as follows:

$\begin{array}{c}\mathrm{F}=\frac{(\mathrm{Q}/2)(\mathrm{Q}/2)}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\\ =\frac{1}{4}\frac{{\mathrm{Q}}^2}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\\ =\frac{1}{4}\frac{(9×{10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}){(6.0×{10}^{−6}\text{C})}^2}{{(3.00×{10}^{−3}\text{m})}^2}\\ =9.0×{10}^3\text{N}\end{array}$

Hence, the value of maximum possible force is $9.0×{10}^3\text{N}$.