91Ó°ÊÓ

The given data is listed as follows,

• The length of the uniformly charged rod is, $l=0.4\mathrm{m}$.
• The charge carried by the rod is, Q=2.2nC.
• The distance at which the electric field’s magnitude is to be found,$r=3\mathrm{cm}.$

The magnitude of the electric field is directly proportional to the charge carried by the field and inversely proportional to the product of the root of the square of the distance of the electric field and half of the length of the field and the distance of the field.

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

The equation of the magnitude of the electric field can be expressed as:

$E=k\frac{Q}{r\sqrt{r^2+{(L/2)}^2}}$

$E=k\frac{Q}{r\sqrt{r^2+{(L/2)}^2}}$

Here, $k$is the electric field constant with value $9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2,\mathrm{Q}$is the charge,$L$is the length of the uniformly charged rod and is the distance of the magnitude of the electric field from the rod’s midpoint.

Substitute all the values in the above equation,

$$\begin{gathered} E=9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×\frac{2.2\mathrm{nC}×{\displaystyle \frac{1×{10}^{-9}\mathrm{C}}{1\mathrm{nC}}}}{\left(3\mathrm{cm}×\frac{1\mathrm{m}}{100\mathrm{cm}}\right)×\sqrt{{\left(3\mathrm{cm}×\frac{1\mathrm{m}}{100\mathrm{cm}}\right)}^2+{\left({\displaystyle \frac{0.4\mathrm{m}}{2}}\right)}^2}} \\ 9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×\frac{2.2×{10}^{-9}\mathrm{C}}{(0.03\mathrm{m})×\sqrt{{(0.03\mathrm{m})}^2+{(0.2\mathrm{m})}^2}} \end{gathered}$$

localid="1656929822624" $$\begin{gathered} =9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×3.63×{10}^{-7}\mathrm{C}/{\mathrm{m}}^2 \\ =3263.48\mathrm{N}/\mathrm{C}. \end{gathered}$$

Thus, the magnitude of the electric field at a location from the midpoint of the rod is $3263.48\mathrm{N}/\mathrm{C}.$