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 4.00 cm from the center of the solid cylinder is approximately 1436.18 N/C.

Step-by-step solution

Recall Gauss's law for electric fields

Gauss's law states that the total electric flux through a closed surface is proportional to the charge enclosed by that surface, given by the equation: \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}\) Where \(\vec{E}\) is the electric field, \(d\vec{A}\) is an infinitesimal area vector, \(Q_{\text{enclosed}}\) is the total charge enclosed by the Gaussian surface, and \(\epsilon_0\) is the vacuum permittivity.

Choose a suitable Gaussian surface

In this case, we will use a cylindrical Gaussian surface, also known as a pillbox. The cylinder should have a radius r (in this case, 4.00 cm) and be centered around the y-axis, as shown in the problem. The height of the cylinder doesn't matter because we are dealing with an infinitely long cylinder.

Determine symmetry

Because the cylinder is infinitely long and has a uniform charge distribution, the electric field will have cylindrical symmetry. This means that the electric field is everywhere perpendicular to the Gaussian surfaces, resulting in the areas of the top and bottom surfaces having no contribution to the total flux through these parts. So, the total electric flux will be calculated only through the curved surface of the Gaussian cylinder.

Calculate the electric flux through the Gaussian surface

Now we calculate the electric flux through the curved surface of the Gaussian cylinder. Since the electric field is everywhere perpendicular to the Gaussian surfaces, the electric flux is simply the product of the electric field E and the surface area A, given by: \(\int \vec{E} \cdot d\vec{A} = EA\) Where A is the surface area of the curved part of the Gaussian cylinder, equal to the circumference of the circle (2πr) multiplied by the height of the cylinder (h). \(A = 2\pi rh\)

Calculate the enclosed charge

Next, we calculate the charge enclosed by the Gaussian cylinder. The charge per unit volume is given as \(\rho\). The volume of the Gaussian cylinder with radius r and height h is given by: \(V = \pi r^2 h\) The total charge enclosed can be calculated as: \(Q_{\text{enclosed}} = \rho V = \rho \pi r^2 h\)

Apply Gauss's law to find the electric field

Now we apply Gauss's law to find the electric field at radius r: \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}\) \(EA = \frac{\rho \pi r^2 h}{\epsilon_0}\) Since we want to calculate the electric field E, we can rearrange the equation and solve for E: \(E = \frac{\rho \pi r^2 h}{\epsilon_0 A}\) Substitute the expression for the surface area A and simplify: \(E = \frac{\rho \pi r^2 h}{\epsilon_0 2 \pi rh}\) \(E = \frac{\rho r}{2 \epsilon_0}\) Now, plug the given values of r and ρ into the equation: \(E = \frac{(6.4 \cdot 10^{-8} \mathrm{C/m}^3)(0.04 \mathrm{m})}{2(8.854 \cdot 10^{-12} \mathrm{C^2/Nm^2})}\) Calculate the electric field: \(E \approx 1436.18 \,\text{N/C}\) The magnitude of the electric field at a radius of 4.00 cm from the center of the solid cylinder is approximately 1436.18 N/C.