Answer:
The magnitude of the net electric field is [tex]6.57\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Charge density [tex]\lambda = 4.54\ \mu C/m[/tex]
Charge density [tex]\lambda' = -2.58\ \mu C/m[/tex]
Distance [tex]y_{1}= 0.384\ m[/tex]
Distance [tex]y_{2}= 0.204\ m[/tex]
We need to calculate the magnitude of the net electric field
Using formula of electric field
[tex]E=E_{1}+E_{2}[/tex]
[tex]E=\dfrac{1}{2\pi\epsilon_{0}}(\dfrac{\lambda}{r}+\dfrac{\lambda'}{r'})[/tex]
Put the value into the formula
[tex]E=\dfrac{1}{2\pi\times8.85\times10^{-12}}(\dfrac{4.54\times10^{-6}}{0.204}+\dfrac{2.58\times10^{-6}}{0.384-0.204})[/tex]
[tex]E=6.57\times10^{5}\ N/C[/tex]
Hence, The magnitude of the net electric field is [tex]6.57\times10^{5}\ N/C[/tex]