Respuesta :
Explanation:
It is given that,
Length of the rod, l = 14 cm = 0.14 m
Charge on the rod, [tex]q=-22\ \mu C=-22\times 10^{-6}\ C[/tex]
We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m
Electric field at the axis of the rod is given by :
[tex]E=\dfrac{\lambda}{2\pi \epsilon_o z}[/tex]
Where
[tex]\lambda[/tex] is the linear charge density of the rod,
[tex]\lambda=\dfrac{q}{l}=\dfrac{-22\times 10^{-6}\ C}{0.14\ m}=-0.00015\ C/m[/tex]
[tex]E=\dfrac{-0.00015}{2\pi\times 8.85\times 10^{-12}\times 0.36}[/tex]
E = -7493170.57 N/C
or
[tex]E=-7.49\times 10^6\ N/C[/tex]
Negative sign shows that the electric field is acting in inwards direction. Hence, this is the required solution.
