91Ó°ÊÓ

Provided:

The Coulomb’s law is $F=\frac{k{q}_1{q}_2}{r^2}$where $q_1$and $q_2$are charged.

$q=3\%,r=5\%,F=2\%$

This method is based on the derivatives of functions whose values must be calculated at certain locations, as the name implies.

Consider a function $\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, find the value of a function $\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$when.

As an example, the derivative of a function$\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$with respect to xwill be employed.

$\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{=}\mathbf{\left(}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right)}$ is Change in y with respect to change in x as$\mathit{d}\mathit{x}\mathbf{⇒}\mathbf{0}$ .

If the value of $x=x\text{'}$from a value of x near it, such that the difference in the two values, dx, is vanishingly small, one can derive the change in the value of the function $y=f\left(\mathrm{x}\right)$corresponding to the change dx in x. In practice, however, the concept of vanishingly small is not possible.

Consider the relative errors for two of the three variables, differentials may be used to calculate the relative error in $g\left(g_{ge}\right)$.

It will, however, need to first extract the relationship's differentials:

Since,

$F=\frac{k{q}_1{q}_2}{r^2}$

Take log on both sides,

$\mathrm{InF}=\mathrm{Ink}+{\mathrm{Inq}}_1+{\mathrm{Inq}}_2-2\mathrm{Inr}$

Here,

$\begin{array}{l}\frac{dF}{F}=\frac{d{q}_1}{q_1}+\frac{d{q}_2}{q_2}-2\frac{dr}{r}\\ \frac{d{q}_2}{q_2}=\frac{dF}{F}-\frac{d{q}_1}{q_1}+2\frac{dr}{r}\end{array}$

In the worst case,$\frac{d{q}_1}{q_1}<0.$

$\frac{d{q}_2}{q_2}=\frac{dF}{F}+\frac{d{q}_1}{q_1}+2\frac{dr}{r}$

Put the values and solve it,

$$\begin{gathered} q_{2re}=0.03+0.02+2×0.05 \\ =0.15 \end{gathered}$$

Hence,

$q_{2re}=15\%$