Answer:
The magnitude of the electric field at a point equidistant from the lines is [tex]4.08\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
[tex]\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})[/tex]
[tex]\theta=\sin^{-1}(\dfrac{1}{4})[/tex]
[tex]\theta=14.47^{\circ}[/tex]
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
[tex]E=\dfrac{2k\lambda}{r}\times2\cos\theat[/tex]
Put the value into the formula
[tex]E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}[/tex]
[tex]E=408094.00\ N/C[/tex]
[tex]E=4.08\times10^{5}\ N/C[/tex]
Hence, The magnitude of the electric field at a point equidistant from the lines is [tex]4.08\times10^{5}\ N/C[/tex]