91Ó°ÊÓ

The given data is listed below as-

• The length of the thin rod is,$L=0.4\mathrm{m}$
• The charge of the thin rod is,$\mathrm{Q}=2.5\mathrm{n}/\mathrm{C}$
• The location of the electric field is,$X=1\mathrm{cm}$

The electric field is a region in which a charged particle is able to exert force on another charged particle.

The concept of the electric field gives the magnitude of the electric field.

The equation of the magnitude of the electric field is expressed as,

$E=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{Q}{\sqrt[x]{x^2+{\left({\displaystyle \frac{L}{2}}\right)}^2}}\right)$

Here,role="math" localid="1656051019749" $\frac{1}{4{\mathrm{Ï€Î\mu }}_0}$is the constant of electric field,$Q$ is the charge of the thin rod, $X$is the location of the electric field and is the length of the thin rod.

For,$\frac{1}{4{\mathrm{Ï€Î\mu }}_0}=8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$, $\mathrm{Q}=2.5\mathrm{nC}$,$\mathrm{x}=1\mathrm{cm}$and$L=0.4\mathrm{m}$

$$\begin{gathered} E=8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×\left(\frac{2.5\mathrm{nC}}{\sqrt[1\mathrm{cm}]{{\left(1\mathrm{cm}\right)}^2+{\left({\displaystyle \frac{0.4\mathrm{m}}{2}}\right)}^2}}\right) \\ =8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\left(\frac{2.5\mathrm{nC}×{\displaystyle \frac{1×{10}^{-9}\mathrm{C}}{1\mathrm{nC}}}}{1\mathrm{cm}×{\displaystyle \frac{1\mathrm{cm}}{100\mathrm{cm}}}\sqrt{{\left(1\mathrm{cm}×{\displaystyle \frac{1\mathrm{cm}}{100\mathrm{cm}}}\right)}^2+{\left({\displaystyle \frac{0.4\mathrm{m}}{2}}\right)}^2}}\right) \\ =8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\left(\frac{2.5×{10}^{-9}\mathrm{C}}{0.01\mathrm{m}\sqrt{{\left(0.01\right)}^2+{\left({\displaystyle \frac{0.04\mathrm{m}}{2}}\right)}^2}}\right) \\ =8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×\left(\frac{2.5×{10}^{-9}\mathrm{C}}{2.002×{10}^{-3}{\mathrm{m}}^2}\right) \end{gathered}$$

Hence, further as,

$$\begin{gathered} E=8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×\left(\frac{2.5×{10}^{-9}\mathrm{C}}{2.002×{10}^{-3}{\mathrm{m}}^2}\right) \\ =8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×1.24×{10}^{-8}\mathrm{C}/{\mathrm{m}}^2 \\ =11220.42\mathrm{N}/\mathrm{C} \\ =1.12×{10}^4\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the magnitude of the electric field is $1.12×{10}^4\mathrm{N}/\mathrm{C}$.