Chapter 24, Problem 24P

(a)

To determine

The electric field at $10.0\text{cm}$ perpendicular to the length of the filament.

Answer to Problem 24P

The electric field at $10.0\text{cm}$ is $16.2\text{MN}/\text{C}$.

Explanation of Solution

Consider the charged filament is coincided with the Gaussian cylinder of length $l$ and radius $r$.

Write the expression for Gauss law.

    ${\displaystyle \int E\cdot da}=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

Here, $E$ is the electric field, $Q_{\text{enc}}$ is the enclosed charge, $a$ is the surface area of the Gaussian cylinder and ${\epsilon }_0$ is the permittivity of free space.

Substitute $\left|E\right|\cdot \left(2\times \pi \times r\times l\right)$ for ${\displaystyle \int E\cdot da}$ and $\lambda \times l$ for $Q_{\text{enc}}$ in the above equation.

    $\begin{array}{l}\left|E\right|\cdot \left(2\times \pi \times r\times l\right)=\frac{\lambda \times l}{{\epsilon }_0}\\ \left|E\right|=\frac{\lambda }{{\epsilon }_0\left(2\times \pi \times r\right)}\\ \left|E\right|=\frac{2\lambda }{{\epsilon }_0\left(4\times \pi \times r\right)}\\ \left|E\right|=k_{\text{e}}\times \frac{2\times \lambda }{r}\end{array}$                                                                                           (I)

Here, $\lambda$ is the charge per unit length, $\left|E\right|$ is the electric field, $r$ is the radius of the Gaussian cylinder and $k_{\text{e}}$ is the coulomb constant.

Conclusion:

Substitute $8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $10.0\text{cm}$ for $r$ and $-90.0\mu \text{C}/\text{m}$ for $\lambda$ in Equation (I) to calculate $\left|E\right|$.

    $\begin{array}{c}E=8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(\frac{2\times \left(-90.0\mu \text{C}/\text{m}\right)\left(\frac{1\times {10}^{-6}\text{C}/\text{m}}{1\mu \text{C}/\text{m}}\right)}{10.0\text{cm}\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)}\right)\\ =8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(\frac{2\times \left(-90.0\times {10}^{-6}\text{C}/\text{m}\right)}{0.1\text{m}}\right)\\ =8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(-1800\text{C}/{\text{m}}^2\times {10}^{-6}\right)\\ =-16.2\times {10}^6\text{N}/\text{C}\left(\frac{1\text{MN}/\text{C}}{1\times {10}^6\text{N}/\text{C}}\right)\end{array}$

Further solve the above equation.

    $\left|E\right|=16.2\text{MN}/\text{C}$

Therefore, the electric field at $10.0\text{cm}$ is $16.2\text{MN}/\text{C}$.

(b)

To determine

The electric field at $20.0\text{cm}$, perpendicular to the filament.

Answer to Problem 24P

The electric field at $20.0\text{cm}$ is $8.09\text{MN}/\text{C}$.

Explanation of Solution

Conclusion:

Substitute $8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $20.0\text{cm}$ for $r$ and $-90.0\mu \text{C}/\text{m}$ for $\lambda$ in Equation (I) to calculate $\left|E\right|$.

    $\begin{array}{c}E=8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(\frac{2\times \left(-90.0\mu \text{C}/\text{m}\right)\left(\frac{1\times {10}^{-6}\text{C}/\text{m}}{1\mu \text{C}/\text{m}}\right)}{20.0\text{cm}\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)}\right)\\ =8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(\frac{2\times \left(-90.0\times {10}^{-6}\text{C}/\text{m}\right)}{0.2\text{m}}\right)\\ =8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(-900\text{C}/{\text{m}}^2\times {10}^{-6}\right)\\ =-8088300\text{N}/\text{C}\times \left(\frac{1\text{MN}/\text{C}}{1\times {10}^6\text{N}/\text{C}}\right)\end{array}$

Further solve the above equation.

    $\left|E\right|=8.09\text{MN}/\text{C}$

Therefore, the electric field at $20.0\text{cm}$ is $8.09\text{MN}/\text{C}$.

(c)

To determine

The electric field at $100\text{cm}$, perpendicular to the filament

Answer to Problem 24P

The electric field at $100\text{cm}$ is $1.62\text{MN}/\text{C}$.

Explanation of Solution

Conclusion:

Substitute $8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $100\text{cm}$ for $r$ and $-90.0\mu \text{C}/\text{m}$ for $\lambda$ in Equation (I) to calculate $\left|E\right|$.

    $\begin{array}{c}E=8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(\frac{2\times \left(-90.0\mu \text{C}/\text{m}\right)\left(\frac{1\times {10}^{-6}\text{C}/\text{m}}{1\mu \text{C}/\text{m}}\right)}{100\text{cm}\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)}\right)\\ =8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(\frac{2\times \left(-90.0\times {10}^{-6}\text{C}/\text{m}\right)}{1\text{m}}\right)\\ =8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(-180\text{C}/{\text{m}}^2\times {10}^{-6}\right)\\ =-1617660\text{N}/\text{C}\left(\frac{1\text{MN}/\text{C}}{1\times {10}^6\text{N}/\text{C}}\right)\end{array}$

Further solve the above equation.

    $\left|E\right|=1.62\text{MN}/\text{C}$

Therefore, the electric field at $100\text{cm}$ is $1.62\text{MN}/\text{C}$.