Respuesta :
(a) The electric field a short distance r from the cylinder will be [tex]\rm E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]
(b) The distance from the surface is 0.10 cm.
What is gauss law?
Gauss' law may be used to calculate the electric field of an infinite line charge with a uniform linear charge density. The electric field has the same amplitude at every point of the cylinder.
(a) The electric field value at a short distance r from the cylinder will be [tex]\rm E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex].
Gauss' law may be used to calculate the electric field of an infinite line charge with a uniform linear charge density. The electric field has the same amplitude at every point of the cylinder.
It is directed outward when considering a Gaussian surface in the form of a cylinder with radius r. The electric flux is just the electric field multiplied by the cylinder's area.
Hence the electric field value at a short distance r from the cylinder will be [tex]\rm E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex].
(b) The distance from the surface is 0.10 cm.
The potential difference is given by;
[tex]\rm \triangle V= \frac{\lambda}{2\pi \epsilon_0} ln\frac{r_2}{r_1} \\\\ \rm \ln\frac{r_2}{r_1} = e^{2\pi \epsilon_0 \times \frac{\triangle V}{\lambda}[/tex]
[tex]\rm r_2= r-1 \frac{\triangle V}{\lambda} \times r_1 \times e^{2\pi \epsilon_0[/tex]
[tex]\rm r_2= 2.40 \times \frac{175}{15 \times 10^{-6}} \times \times e^{2\times 3.14 \times 8.85 \times 10^{-12} \\\\[/tex]
[tex]\rm r_2=2.410\ m[/tex]
The difference in the radius is the distance above the surface.
d = r₂-r₁
d=2.50-2.410
d=0.09 m
Hence the distance from the surface is 0.10 cm.
To learn more about the gauss law refer to the link;
https://brainly.com/question/2854215
