Explanation:
It is given that,
Distance, r = 3.5 m
Electric field due to an infinite wall of charges, E = 125 N/C
We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :
[tex]E=\dfrac{\lambda}{2\pi \epsilon_o r}[/tex]
It is clear that the electric field is inversely proportional to the distance. So,
[tex]\dfrac{E}{E'}=\dfrac{r'}{r}[/tex]
[tex]E'=\dfrac{Er}{r'}[/tex]
[tex]E'=\dfrac{125\times 3.5}{1.5}[/tex]
E' = 291.67 N/C
So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.
