Question

A thin glass rod of length 80 cmis rubbed all over with wool and acquires a charge of 60 nC, distributed uniformly over its surface. Calculate the magnitude of the electric field due to the rod at a location 7 cmfrom the midpoint of the rod. Do the calculation two ways, first using the exact equation for a rod of any length, and second using the approximate equation for a long rod.

Short answer

The magnitude of the electric field for a rod of any length is1900.07 N/C. The magnitude of the electric field for a rod of any length is1928.57 N/C.

Step-by-step solution

Identification of the given data

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.

Significance of the electric field

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.

Determination of the magnitude of the electric field for a rod of any length

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

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

Determination of the magnitude of the electric field for a long rod

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.

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

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