91Ó°ÊÓ

1. The two charged particles are on the x-axis with separation, L .
2. The value of the ratio,$\frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}=4$
3. The x-axis scale is set as:${\mathrm{x}}_{\mathrm{s}}=30.0\mathrm{cm}$
4. Particle 2 has a charge of$-{\mathrm{q}}_2=-3\mathrm{e}$

Using the concept of the electric field at a given point, we can get the value of an individual electric field by a charge. Again for the maximum value of the net field, we can differentiate the electric field equation for getting the value of x. Now, substituting the value of x, we can get the value of the required electric field.

Formulae:

The magnitude of the electric field, $E=\frac{q}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{R}}^2}\hat{R}$ (1)

where R = The distance of field point from the charge, and q = charge of the particle

According to the superposition principle, the electric field at a point due to more than one charge,

$$\begin{gathered} \stackrel{→}{\mathrm{E}}=\underset{\mathrm{i}-1}{\overset{\mathrm{n}}{∑}}\stackrel{→}{{\mathrm{E}}_{\mathrm{i}}} \\ =\underset{\mathrm{i}-1}{\overset{\mathrm{n}}{∑}}\frac{{\mathrm{q}}_{\mathrm{i}}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_{\mathrm{i}}^2}{\hat{\mathrm{r}}}_{\mathrm{i}} \end{gathered}$$ (2)

For it to be possible for the net field to vanish at some x > 0, the two individual fields (caused by ${\mathrm{q}}_1$and ${\mathrm{q}}_2$) must point in opposite directions for x > 0. They are therefore oppositely charged considering their positions. Further, since the net field points more strongly leftward for the small positive x (where it is very close to ${\mathrm{q}}_2$), then we conclude that localid="1657282744540" ${\mathrm{q}}_2$is the negative-valued charge. Thus, ${\mathrm{q}}_1$is a positive-valued charge.

From the given ratio, we can now considered for getting a maximum electric field that

${\mathrm{q}}_1=4\mathrm{e}\mathrm{and}{\mathrm{q}}_2=\mathrm{e}$

Thus using equation (1), we can find the individual fields, and substituting this into equation (2), we can get the net electric as:

$$\begin{gathered} E_{net}=E_1+E_2 \\ =\frac{4}{4{\mathrm{Ï€Î\mu }}_0{(\mathrm{L}+\mathrm{x})}^2}-\frac{e}{4{\mathrm{Ï€Î\mu }}_0{\left(x\right)}^2}...................\left(3\right) \end{gathered}$$

Setting ${\mathrm{E}}_{\mathrm{net}}=0$at (see graph) x = 20 cm , the graph immediately leads to To get the maximum value of the electric field, we can differentiate the above equation for x, and equating it to zero, we can get the value of x as:

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{4\mathrm{e}}{4{\mathrm{Ï€Î\mu }}_0{(\mathrm{L}+\mathrm{x})}^2}-\frac{\mathrm{e}}{4{\mathrm{Ï€Î\mu }}_0{\left(\mathrm{x}\right)}^2}\right]=0 \\ \mathrm{x}=\left(\frac{2}{3}\sqrt[3]{2}+\frac{1}{3}\sqrt[3]{4}+\frac{1}{3}\right)\mathrm{L} \\ \mathrm{x}=1.70(20\mathrm{cm}) \\ \mathrm{x}=34\mathrm{cm} \end{gathered}$$

Hence, the value of x is 34 cm

Substituting the given values in equation (3), we can the maximum electric field for the charge of particle 2,as follows:

$$\begin{gathered} \mathrm{E}=\frac{4\mathrm{e}}{4{\mathrm{Ï€Î\mu }}_0(\mathrm{L}+\mathrm{x})}-\frac{3\mathrm{e}}{4{\mathrm{Ï€Î\mu }}_0{\left(\mathrm{x}\right)}^2} \\ =9×{10}^9\mathrm{N}.{\mathrm{m}}^2.{\mathrm{C}}^{-2}×1.6×{10}^{-19}\mathrm{C}\left[\left(\frac{4}{{(0.34\mathrm{m}+0.20\mathrm{m})}^2}\right)-\left(\frac{3}{{(0.20\mathrm{m})}^2}\right)\right] \\ =2.2×{10}^{-8}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $2.2×{10}^{-8}\mathrm{N}/\mathrm{C}$