Question

An infinitely long, solid cylinder of radius \(R=9.00 \mathrm{~cm},\) with a uniform charge per unit of volume of \(\rho=6.40 \cdot 10^{-8} \mathrm{C} / \mathrm{m}^{3},\) is centered about the \(y\) -axis. Find the magnitude of the electric field at a radius \(r=4.00 \mathrm{~cm}\) from the center of this cylinder.

Short answer

Answer: The magnitude of the electric field at a distance of 0.04 m from the center of the cylinder is approximately \(1.44 \times 10^5 \, N/C\).

Step-by-step solution

Applying Gauss's Law to find the electric field

Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by the Gaussian surface divided by the vacuum permittivity. Mathematically, Gauss's law is given by: $$ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} $$ Here, \(\oint \vec{E} \cdot d\vec{A}\) is the electric flux through the surface, \(Q_{enc}\) is the charge enclosed by the Gaussian surface, and \(\epsilon_0\) is the vacuum permittivity.

Setting up the Gaussian surface and finding the charge

To apply Gauss's law, we need a Gaussian surface. Since we are dealing with a cylindrical symmetry, let's choose a cylindrical Gaussian surface with radius r and height h. The charge enclosed by this Gaussian surface can be obtained by multiplying the volumetric charge density 𝜌 by the volume of the cylinder enclosed: $$ Q_{enc} = \rho \cdot V = \rho \cdot (\pi r^2 h) $$

Calculating the electric flux through the Gaussian surface

Now, let's calculate the electric flux through the Gaussian surface. Since the electric field is uniform and parallel to the normal of the Gaussian surface, the electric flux can be calculated as: $$ \oint \vec{E} \cdot d\vec{A} = E \oint dA $$ The total area of the Gaussian surface's lateral side is given by \(A = 2\pi r h\), so $$ E \oint dA = EA = E(2\pi rh) $$

Equating flux and charge to find the electric field

Now, we can equate the electric flux and charge from Gauss's law: $$ E(2\pi rh) = \frac{\rho \cdot (\pi r^2 h)}{\epsilon_0} $$ Solve for E: $$ E = \frac{\rho r}{2\epsilon_0} $$

Substituting values and calculating the electric field

Now substitute the given values, \(\rho = 6.40 \times 10^{-8} \, C/m^3\), \(r = 0.04 \, m\), and \(\epsilon_0 = 8.85 \times 10^{-12} \, C^2/Nm^2\), in the equation and calculate the electric field: $$ E = \frac{(6.40 \times 10^{-8} \, C/m^3) (0.04 \, m)}{2(8.85 \times 10^{-12} \, C^2/Nm^2)} $$ After calculating, we get: $$ E \approx 1.44 \times 10^5 \, N/C $$ So, the magnitude of the electric field at a radius of \(r = 4.00 \, cm\) from the center of the cylinder is \(1.44 \times 10^5 \, N/C\).