(3). a very long uniform line of charge has charge per unit length ( 4.80 mu mathrm{c} / mathrm{m} ) and lies along the ( x )-axis. a second long uniform line of charge has charge per unit length ( -2.40 mu mathrm{c} / mathrm{m} ) and is parallel to the ( x )-axis at ( y=0.400 mathrm{~m} ). what is the net electric field (magnitude and direction) at the following points on the ( y )-axis: (a) ( y=0.200 mathrm{~m} ) and (b) ( y=0.600 mathrm{~m} )?

Question

Accepted Answer

# Explanation:
## Step1: Recall the formula for electric field due to a long line charge
The electric field due to a long line charge with charge per unit length $\lambda$ at a distance $r$ from the line charge is given by $E=\frac{\lambda}{2\pi\epsilon_0 r}$, where $\epsilon_0 = 8.85\times10^{-12}\ C^{2}/N\cdot m^{2}$

## Step2: Calculate the electric field at $y = 0.200\ m$
- For the line charge along the $x -$axis ($\lambda_1=4.80\times 10^{-6}\ C/m$), the distance $r_1 = 0.200\ m$
$E_1=\frac{\lambda_1}{2\pi\epsilon_0 r_1}=\frac{4.80\times 10^{-6}}{2\pi\times8.85\times 10^{-12}\times0.200}$
$$
\begin{align*}
E_1&=\frac{4.80\times 10^{-6}}{2\pi\times8.85\times 10^{-12}\times0.200}\
&=\frac{4.80\times 10^{-6}}{1.11\times10^{-11}}\
& = 4.32\times10^{5}\ N/C\ (upwards, since $\lambda_1>0$)
\end{align*}
$$
- For the line charge at $y = 0.400\ m$ ($\lambda_2=- 2.40\times 10^{-6}\ C/m$), the distance $r_2=0.400 - 0.200=0.200\ m$
$E_2=\frac{\lambda_2}{2\pi\epsilon_0 r_2}=\frac{-2.40\times 10^{-6}}{2\pi\times8.85\times 10^{-12}\times0.200}$
$$
\begin{align*}
E_2&=\frac{-2.40\times 10^{-6}}{1.11\times10^{-11}}\
&=- 2.16\times10^{5}\ N/C\ (upwards, since $\lambda_2<0$, the field is towards the line charge)
\end{align*}
$$
- Net electric field $E = E_1+E_2=(4.32\times 10^{5}-2.16\times 10^{5})\ N/C = 2.16\times10^{5}\ N/C$ (upwards)

## Step3: Calculate the electric field at $y = 0.600\ m$
- For the line charge along the $x -$axis ($\lambda_1 = 4.80\times 10^{-6}\ C/m$), the distance $r_1=0.600\ m$
$E_1=\frac{\lambda_1}{2\pi\epsilon_0 r_1}=\frac{4.80\times 10^{-6}}{2\pi\times8.85\times 10^{-12}\times0.600}$
$$
\begin{align*}
E_1&=\frac{4.80\times 10^{-6}}{3.33\times10^{-11}}\
&=1.44\times 10^{5}\ N/C\ (upwards)
\end{align*}
$$
- For the line charge at $y = 0.400\ m$ ($\lambda_2=-2.40\times 10^{-6}\ C/m$), the distance $r_2=0.600 - 0.400 = 0.200\ m$
$E_2=\frac{\lambda_2}{2\pi\epsilon_0 r_2}=\frac{-2.40\times 10^{-6}}{2\pi\times8.85\times 10^{-12}\times0.200}$
$$
\begin{align*}
E_2&=\frac{-2.40\times 10^{-6}}{1.11\times10^{-11}}\
&=-2.16\times 10^{5}\ N/C\ (downwards, since $\lambda_2<0$, the field is towards the line charge)
\end{align*}
$$
- Net electric field $E=E_1 + E_2=(1.44\times 10^{5}-2.16\times 10^{5})\ N/C=- 7.2\times10^{4}\ N/C$ (downwards)

# Answer:
(a) The magnitude of the net electric field at $y = 0.200\ m$ is $2.16\times 10^{5}\ N/C$ and the direction is upwards.

(b) The magnitude of the net electric field at $y = 0.600\ m$ is $7.2\times 10^{4}\ N/C$ and the direction is downwards.

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the formula for electric field due to a long line charge

Step2: Calculate the electric field at \(y = 0.200\ m\)

Step3: Calculate the electric field at \(y = 0.600\ m\)

Answer: