Chapter 16, Problem 74E

(a)

To determine

The magnitude and direction of the field due to the charges.

Answer to Problem 74E

The magnitude of net electric field above the surface of the earth is $\underset{\_}{\left(45000\text{N}/\text{C}\right)}$ and is directed $\underset{\_}{\text{downward towards the ground}}$.

Explanation of Solution

“The electric force per unit charge is defined as the electric field”.

The expression for the electric field due to a point charge is as follows:

  $\overrightarrow{E}=\frac{kQ}{r^2}\widehat{r}$        (I)

Here, $\overrightarrow{E}$ is the electric field, $Q$ is the charge, $k$ is the Coulomb’s constant, and $r$ is the distance.

Analogous to equation (I), the electric field at the surface of the earth just below the cloud due to the charge $+50\text{C}$ can be written as,

  ${\overrightarrow{E}}_1=\frac{k{Q}_1}{r_1^2}\widehat{j}$        (II)

Here, ${\overrightarrow{E}}_1$ is the electric field due to the charge $+50\text{C}$ and $r_1$ is the distance between the charge 1 and the ground.

Analogous to equation (I), the electric field at the surface of the earth just below the cloud due to the charge $-20\text{C}$ can be written as,

  ${\overrightarrow{E}}_2=\frac{k{Q}_2}{r_2^2}\left(-\widehat{j}\right)$        (III)

Here, ${\overrightarrow{E}}_2$ is the electric field due to the charge $-20\text{C}$ and $r_2$ is the distance between the charge 2 and the ground.

Write the expression for the net electric field.

  ${\overrightarrow{E}}_{\text{net}}={\overrightarrow{E}}_1-{\overrightarrow{E}}_2$        (IV)

Here, ${\overrightarrow{E}}_{\text{net}}$ is the net electric field.

Use equation (II) and (III) in equation (IV).

  ${\overrightarrow{E}}_{\text{net}}=\frac{k{Q}_1}{r_1^2}\widehat{j}-\frac{k{Q}_2}{r_2^2}\left(-\widehat{j}\right)$

Conclusion:

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $+50\text{C}$ for $Q_1$, $10\text{km}$ for $r_1$, $-20\text{C}$ for $Q_2$, and $2\text{km}$ for $r_2$.

  $\begin{array}{c}{\overrightarrow{E}}_{\text{net}}=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(+50\text{C}\right)}{{\left(10\text{km}\right)}^2}\widehat{j}-\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(-20\text{C}\right)}{{\left(2\text{km}\right)}^2}\left(-\widehat{j}\right)\\ =\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(+50\text{C}\right)}{{\left(10\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}^2}\widehat{j}-\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(-20\text{C}\right)}{{\left(2\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}^2}\left(-\widehat{j}\right)\\ ==\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(+50\text{C}\right)}{{\left(10\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}^2}\widehat{j}-\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(20\text{C}\right)}{{\left(2\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}^2}\widehat{j}\\ =\left(4500\text{N}/\text{C}\right)\widehat{j}-\left(45000\text{N}/\text{C}\right)\widehat{j}\\ =\left(45000\text{N}/\text{C}\right)\left(-\widehat{j}\right)\end{array}$

The direction of the electric field due to both the charges at the surface of the earth is downward.

Therefore, the magnitude of net electric field above the surface of the earth is $\left(45000\text{N}/\text{C}\right)$ and is directed downward towards the ground.

(b)

To determine

The locations and values of the image charges corresponding to the $+50\text{C}$ and $-20\text{C}$ charges.

Answer to Problem 74E

The locations and values of the image charges corresponding to the $+50\text{C}$ and $-20\text{C}$ charges is shown in figure 1.

Explanation of Solution

The image charge for $+50\text{C}$ is $-50\text{C}$ and the image charge for $-20\text{C}$ is $+20\text{C}$ and they are located at the same distance below the surface as of the same distance they have located above the surface.

Conclusion:

The image charge for $+50\text{C}$ is $-50\text{C}$ and the image charge for $-20\text{C}$ is $+20\text{C}$ is shown below.

(c)

To determine

The total electric field due to all the charges at the surface of earth just below the cloud.

Answer to Problem 74E

The total electric field due to the charges at the surface of the earth just below the cloud is $\underset{\_}{\left(-20250\text{N}/\text{C}\right)\widehat{j}}$.

Explanation of Solution

“The electric force per unit charge is defined as the electric field”.

The expression for the electric field due to a point charge is as follows:

  $\overrightarrow{E}=\frac{kQ}{r^2}\widehat{r}$

Here, $\overrightarrow{E}$ is the electric field, $Q$ is the charge, $k$ is the Coulomb’s constant, and $r$ is the distance.

Write the expression for the electric field due to the charges $-20\text{C}$ and $20\text{C}$ is,

  $E_1^{\text{'}}=\frac{k{Q}_{+20\text{c}}}{d_1^2}(-j)+\frac{k{Q}_{-20\text{c}}}{d_1^2}(-j)$        (V)

Here, $E_1^{\text{'}}$ is the electric field due to the charges $+20\text{C}$ and $-20\text{C}$ and $d_1$ is the distance between the charges between $+20\text{C}$ and $-20\text{C}$.

Write the expression for the electric field due to the charges $-50\text{C}$ and $50\text{C}$ is,

  $E_2^{\text{'}}=\frac{k{Q}_{+50\text{c}}}{d_2^2}(j)+\frac{k{Q}_{-50\text{c}}}{d_2^2}(j)$        (VI)

Here, $E_2^{\text{'}}$ is the electric field due to the charges $+50\text{C}$ and $-50\text{C}$ and $d_2$ is the distance between the charges between $+20\text{C}$ and $-20\text{C}$.

Write the expression for the total electric field.

  $E_{\text{total}}=E_1^{\text{'}}+E_2^{\text{'}}$        (VII)

Here, $E_{\text{total}}$ is the net electric field.

Use equation (V) and (VI) in equation (VII).

  $E_{\text{total}}=\frac{k{Q}_{+20\text{c}}}{d_1^2}(-j)+\frac{k{Q}_{-20\text{c}}}{d_1^2}(-j)+\frac{k{Q}_{+50\text{c}}}{d_2^2}(j)+\frac{k{Q}_{-50\text{c}}}{d_2^2}(j)$

Conclusion:

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $50\text{C}$ for $Q_{+50\text{C}}$, $50\text{C}$ for $Q_{-50\text{C}}$ $20\text{km}$ for $d_1$, $+20\text{C}$ for $Q_{+20\text{C}}$, $20\text{C}$ for $Q_{-20\text{C}}$ and $4\text{km}$ for $d_2$.

  $\begin{array}{c}{E}_{\text{total}}=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(20\text{C}\right)}{{\left(4\text{km}\right)}^2}(-\widehat{j})+\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(20\text{C}\right)}{{\left(4\text{km}\right)}^2}(-\widehat{j})+\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(50\text{C}\right)}{{\left(20\text{km}\right)}^2}(\widehat{j})+\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(50\text{C}\right)}{{\left(20\text{km}\right)}^2}(\widehat{j})\\ =\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(20\text{C}\right)}{{\left(4\text{km}\left(\frac{{10}^3\text{m}}{1\text{km}}\right)\right)}^2}(-\widehat{j})+\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(20\text{C}\right)}{{\left(4\text{km}\left(\frac{{10}^3\text{m}}{1\text{km}}\right)\right)}^2}(-\widehat{j})+\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(50\text{C}\right)}{{\left(20\text{km}\left(\frac{{10}^3\text{m}}{1\text{km}}\right)\right)}^2}(\widehat{j})+\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(50\text{C}\right)}{{\left(20\text{km}\left(\frac{{10}^3\text{m}}{1\text{km}}\right)\right)}^2}(\widehat{j})\\ =\left(-22500\text{N}/\text{C}\right)\widehat{j}+\left(2250\text{N}/\text{C}\right)\widehat{j}\\ =\left(-20250\text{N}/\text{C}\right)\widehat{j}\end{array}$

Therefore, the total electric field due to the charges at the surface of the earth just below the cloud is $\left(-20250\text{N}/\text{C}\right)\widehat{j}$.

(d)

To determine

Whether the lightning from this cloud strike the ground or not.

Answer to Problem 74E

No, the lightning will not strike the ground.

Explanation of Solution

The total electric field due to all the charges is less than the electric field generated by the air as a conductor. Since the electric field due to all the charges is less than the electric field generated by the air as a conductor, the cloud lightning does not strike the ground.

Conclusion:

Since the electric field due to all the charges is less than the electric field generated by the air as a conductor, the cloud lightning does not strike the ground.

Therefore, the lightning from the cloud does not strike the ground.