Charge of uniform density (80 nC/m³) is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical surfaces of radii 1.0 mm and 3.0 mm. Determine the magnitude of the electric field at a point which is 2.0 mm from the symmetry axis. A) 7.9 N/C B) 9.0 N/C C) 5.9 N/C D) 6.8 N/C E) 18 N/C

Question

Quizplus · Accepted Answer

Using Gauss's law, we can find the electric field at a distance of 2.0 mm from the symmetry axis as follows:
$\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$
Since the electric field is uniform and parallel to the surface of the cylinder, the dot product simplifies to $E\oint dA = EA$, and the electric field is taken out of the integral. We choose a Gaussian surface in the form of a cylinder with a radius of 2.0 mm and a height of L:
$\oint \vec{E} \cdot d\vec{A} = E(2\pi rL) = \frac{Q}{\epsilon_0}$
The total charge enclosed by this surface is given by:
$Q = \lambda (\pi r_2^2 - \pi r_1^2)L$
where $\lambda$ is the charge density given in the problem, and $r_1$ and $r_2$ are the inner and outer radii of the cylinder, respectively. Substituting these values and solving for the electric field, we obtain:
$E = \frac{\lambda}{2\pi\epsilon_0}\ln(\frac{r_2}{r_1})$
$E = \frac{(80 	imes 10^{-9} 	ext{C/m})(1 	ext{mm})}{2\pi(8.854 	imes 10^{-12} 	ext{C}^2/	ext{N}\cdot	ext{m}^2)}\ln(\frac{3 	ext{mm}}{1 	ext{mm}})$
$E = 6.8 	ext{N/C}$
Therefore, the answer is D).

Charge of Uniform Density (80 NC/m3) Is Distributed Throughout a Hollow

Charge of Uniform Density (80 NC/m³) Is Distributed Throughout a Hollow