91Ó°ÊÓ

Given data;

Two positive charge are place at$\mathrm{x}=\mathrm{a}$ and$\mathrm{x}=-\mathrm{a}$

Formula;

The electric field due to charge

$\mathbf{E}\mathbf{=}\frac{\mathbf{Kq}}{{\mathbf{a}}^{\mathbf{2}}}\hat{\mathbf{i}}$ ........... (1)

(a)

Put$\mathrm{x}=\mathrm{a}$ in equation (1)

${\mathrm{E}}_1=\frac{\mathrm{Kq}}{{\mathrm{a}}^2}\mathrm{i}$

Put$\mathrm{x}=-\mathrm{a}$ in equation (1)

${\mathrm{E}}_2=−\frac{\mathrm{Kq}}{{\mathrm{a}}^2}\mathrm{i}$

Total electric filed

$$\begin{gathered} \mathrm{E}={\mathrm{E}}_1+{\mathrm{E}}_2 \\ =\frac{\mathrm{Kq}}{{\mathrm{a}}^2}\mathrm{i}−\frac{\mathrm{Kq}}{{\mathrm{a}}^2}\mathrm{i} \\ =0 \end{gathered}$$

Hence, the magnitude and direction of the $\mathrm{E}=0$

For$\mathrm{x}=\mathrm{a}$ and $\mathrm{x}=-\mathrm{a}$, the electric field

$$\begin{gathered} {\mathrm{E}}_1=\frac{\mathrm{q}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(x+a{)}^2} \\ {\mathrm{E}}_2=−\frac{\mathrm{q}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(x−a{)}^2} \end{gathered}$$

Total electric field

$$\begin{gathered} \mathrm{E}={\mathrm{E}}_1+{\mathrm{E}}_2 \\ =\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0(\mathrm{x}+\mathrm{a}{)}^2}−\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0(\mathrm{x}−\mathrm{a}{)}^2} \\ =\left(\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0}\right)\left[\frac{(\mathrm{x}−\mathrm{a}{)}^2+(\mathrm{x}+\mathrm{a}{)}^2}{\left(\right(\mathrm{x}+\mathrm{a}\left)\right(\mathrm{x}−\mathrm{a}){)}^2}\right] \\ =\left(\frac{\mathrm{q}}{{\mathrm{Ï€Î\mu }}_0}\right)\left(\frac{−\mathrm{ax}}{{\left({\mathrm{x}}^2−{\mathrm{a}}^2\right)}^2}\right) \end{gathered}$$

Hence,$\left|x\right|<a:E=\left(\frac{q}{\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\left(\frac{−ax}{{\left(x^2−a^2\right)}^2}\right)$

For$\mathrm{x}=\mathrm{a}$ and $\mathrm{x}=-\mathrm{a}$, the electric field

$$\begin{gathered} {\mathrm{E}}_1=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(x−a{)}^2} \\ \begin{array}{rl}{E}_2& =−\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(x+a{)}^2}\end{array} \end{gathered}$$

Total electric field

$$\begin{gathered} \begin{array}{rl}E& =E_1+E_2\end{array} \\ =\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(x−a{)}^2}−\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(x+a{)}^2} \\ =\left(\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\left[\frac{(x−a{)}^2+(x+a{)}^2}{\left(\right(x+a\left)\right(x−a){)}^2}\right] \\ =\left(\frac{q}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\left(\frac{x^2+a^2}{{\left(x^2−a^2\right)}^2}\right) \end{gathered}$$

Hence,$\left|x\right|>a:E\left(\frac{q}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\left(\frac{x^2+a^2}{{\left(x^2−a^2\right)}^2}\right)$