Chapter 10, Problem 1P

a. Find the electric field strength at a point 1 m from a charge of 4 $\text{μC}$.

b. Find the electric field strength at a point 1 mm [1/1000 the distance of part (a)] from the same charge as part (a) and compare results.

To determine

(a)

The electric field strength at a point $1\text{ m}$ from a charge of $4\mu \text{C}$.

Answer to Problem 1P

The electric field strength at a point $1\text{ m}$ from a charge of $4\mu \text{C}$ is $35.97\times {10}^3\text{ N/C}$.

Explanation of Solution

Given:

The distance is $1\text{ m}$.

The charge is $4\mu \text{C}$.

Concept used:

Expression for the electric filed due to a point charge.

$E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$ ..... (1)

Here, $\overrightarrow{E}$ is electric field, $Q$ is the charge and ${\epsilon }_0$ is the permittivity of free space.

Calculation:

Substitute $4\mu \text{C}$ for $Q$, $8.85\times {10}^{-12}{\text{ C}}^2{\text{/Nm}}^2$ for ${\epsilon }_0$ and $1\text{ m}$ for $r$ in equation (1).

  $\begin{array}{c}E=\frac{4\mu \text{C}}{4\pi \left(8.85\times {10}^{-12}{\text{ C}}^2{\text{/Nm}}^2\right){\left(1\text{ m}\right)}^2}\\ =\frac{4\mu \text{C}\left(\frac{1\text{ C}}{{10}^6\mu \text{C}}\right)}{4\pi \left(8.85\times {10}^{-12}{\text{ C}}^2{\text{/Nm}}^2\right){\left(1\text{ m}\right)}^2}\\ =\frac{4\times {10}^{-6}\text{ C}}{4\pi \left(8.85\times {10}^{-12}{\text{ C}}^2{\text{/Nm}}^2\right){\left(1\text{ m}\right)}^2}\\ =3.59\times {10}^4\text{ N/C}\end{array}$

Conclusion:

Thus, the electric field strength at a point $1\text{ m}$ from a charge of $4\mu \text{C}$ is $35.97\times {10}^3\text{ N/C}$.

To determine

(b)

The electric field strength at a point $1\text{ mm}$ from a charge of $4\mu \text{C}$. Also, compare the results with the answer obtained in part (a).

Answer to Problem 1P

The electric field strength at point $1\text{ mm}$ is $3.59\times {10}^{10}\text{ N/C}$ and the electric field strength at point $1\text{ mm}$ is ${10}^6$ times greater than the electric field strength at a point $1\text{ m}$ from a charge of $4\mu \text{C}$.

Explanation of Solution

Given:

The distance is $1\text{ mm}$.

The charge is $4\mu \text{C}$.

Calculation:

Substitute $4\mu \text{C}$ for $Q$, $8.85\times {10}^{-12}{\text{ C}}^2{\text{/Nm}}^2$ for ${\epsilon }_0$ and $1\text{ mm}$ for $r$ in equation (1).

  $\begin{array}{c}E=\frac{4\mu \text{C}}{4\pi \left(8.85\times {10}^{-12}{\text{ C}}^2{\text{/Nm}}^2\right){\left(1\text{ mm}\right)}^2}\\ =\frac{4\mu \text{C}\left(\frac{1\text{ C}}{{10}^6\mu \text{C}}\right)}{4\pi \left(8.85\times {10}^{-12}{\text{ C}}^2{\text{/Nm}}^2\right){\left(1\text{ mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)\right)}^2}\\ =\frac{4\times {10}^{-6}\text{ C}}{4\pi \left(8.85\times {10}^{-12}{\text{ C}}^2{\text{/Nm}}^2\right){\left({10}^{-3}\text{ m}\right)}^2}\\ =3.59\times {10}^{10}\text{ N/C}\end{array}$

Therefore, the electric field strength at a point $1\text{ mm}$ is ${10}^6$ times greater than the electric field strength at a point $1\text{ m}$ from a charge of $4\mu \text{C}$

Conclusion:

Thus, the electric field strength at point $1\text{ mm}$ is $3.59\times {10}^{10}\text{ N/C}$ and the electric field strength at point $1\text{ mm}$ is ${10}^6$ times greater than the electric field strength at point $1\text{ m}$ from a charge of $4\mu \text{C}$.