91Ó°ÊÓ

The given data is listed below as:

• The length of the thin glass is, $L=80\mathrm{cm}$.
• The thin glass acquires a charge of $Q=60\mathrm{nC}$.
• The electric field is at the distance of $r=7\mathrm{cm}$from the rod’s midpoint.

The electric field’s magnitude shows a directly proportional relationship to the charge, and it is also inversely related to the particular distance square of the thin glass from the rod.

The equation of the magnitude of the electric field for a rod of any length is expressed as:

$E=k\frac{Q}{\sqrt[r]{r^2+{\left({\displaystyle \frac{L}{2}}\right)}^2}}$

Here,$E$ is the magnitude of the electric field, $k$is the electric field, constant, $Q$is the amount of charge, $r$is the location of the electric field from the rod’s midpoint and $L$is the length of the electric field.

Substitute the values in the above equation.

role="math" localid="1656932557411" $$\begin{gathered} E=9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2×\frac{60\mathrm{nC}\left({\displaystyle \frac{1×{10}^{-9}\mathrm{C}}{1\mathrm{nC}}}\right)}{\left(7\mathrm{cm}\left({\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}\right)\right)×\sqrt{{\left(7\mathrm{cm}\left({\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}\right)\right)}^2+{\left({\displaystyle \frac{\left(80\mathrm{cm}\left({\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}\right)\right)}{\begin{array}{c}2\\ \end{array}}}\right)}^2}} \\ =9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2×\frac{6×{10}^{-9}\mathrm{C}}{\left(0.07\mathrm{m}\right)×\sqrt{{\left(0.07\mathrm{m}\right)}^2+{\left({\displaystyle \frac{\left(0.8\mathrm{m}\right)}{2}}\right)}^2}} \\ =\frac{54\mathrm{N}·{\mathrm{m}}^2/\mathrm{C}}{\left(0.07\mathrm{m}\right)×0.406\mathrm{m}} \\ =1900.07\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, magnitude of the electric field for a rod of any length is $1900.07\mathrm{N}/\mathrm{C}$.

The equation of the magnitude of the electric field for a long rod where the rod’s length is much greater than the radial distance is expressed as:

$E=k\frac{2Q}{rL}$

Here,$E$is the magnitude of the electric field, $k$is the electric field, constant, $Q$is the amount of charge, $r$is the location of the electric field from the rod’s midpoint and$L$ is the length of the electric field.

Substitute the values in the above equation.

$$\begin{gathered} E=9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2×\frac{2×60\mathrm{nC}\left({\displaystyle \frac{1×{10}^{-9}\mathrm{C}}{1\mathrm{nC}}}\right)}{\left(7\mathrm{cm}\left({\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}\right)\right)×\left(80cm\left({\displaystyle \frac{1m}{100cm}}\right)\right)} \\ =9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2×\frac{2×6×{10}^{-9}\mathrm{C}}{\left(0.07\mathrm{m}\right)×\left(0.8\mathrm{m}\right)} \\ =\frac{2×54\mathrm{N}·{\mathrm{m}}^2/\mathrm{C}}{\left(0.07\mathrm{m}\right)×0.8\mathrm{m}} \\ =1928.57\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, magnitude of the electric field for a rod of any length is $1928.57\mathrm{N}/\mathrm{C}$.