Answer:
Distance of the point where electric filed is 2.45 N/C is 1.06 m
Explanation:
We have given charge per unit length, that is liner charge density [tex]\lambda =1.44\times 10^{-10}C/m[/tex]
Electric field E = 2.45 N/C
We have to find the distance at which electric field is 2.45 N/C
We know that electric field due to linear charge is equal to
[tex]E=\frac{\lambda }{2\pi \epsilon _0r}[/tex], here [tex]\lambda[/tex] is linear charge density and r is distance of the point where we have to find the electric field
So [tex]2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}[/tex]
r = 1.06 m
So distance of the point where electric filed is 2.45 N/C is 1.06 m