Question

The cylindrical surface \(\rho=8 \mathrm{~cm}\) contains the surface charge density, \(\rho_{S}=\) \(5 e^{-20|z|} \mathrm{nC} / \mathrm{m}^{2} .(a)\) What is the total amount of charge present? \((b)\) How much electric flux leaves the surface \(\rho=8 \mathrm{~cm}, 1 \mathrm{~cm}

Short answer

\(Φ_E = 5 (8\,\mathrm{cm}) k \left[ \frac{1}{20} (e^{-80} - e^{-20}) \right] (60^\circ)\) \(Φ_E = 40 k \left[ \frac{1}{20} (e^{-80} - e^{-20}) \right] (60^\circ)\) Thus, the electric flux leaving the surface within the specified limits for z and φ is given by \(40 k \left[ \frac{1}{20} (e^{-80} - e^{-20}) \right] (60^\circ)\).

Step-by-step solution

Surface Charge Density Expression

First, we need to write down the surface charge density expression given in the problem. The surface charge density is given as: \(\rho_S = 5 e^{-20|z|} \mathrm{nC/m^2}\)

Calculate Total Charge Present

To find the total charge present on the cylindrical surface, we need to integrate the surface charge density over the entire surface. For a cylindrical surface, the surface area dA can be written as: \(dA = \rho d\phi dz\) The total charge can be calculated by integrating over the entire surface area: \(Q = \int\int \rho_S dA = \int\int 5 e^{-20|z|} d(\rho d\phi dz)\) Since \(\rho\) is constant (8 cm), we integrate with respect to \(\phi\) from 0 to \(2\pi\), and z from \(-\infty\) to \(\infty\). \(Q = 5\rho \int_{0}^{2\pi} d\phi \int_{-\infty}^{\infty} e^{-20|z|} dz\) Now, let's separate the integral into two parts: \(Q = 5\rho \int_{0}^{2\pi} d\phi \left[\int_{-\infty}^{0} e^{20z} dz + \int_{0}^{\infty} e^{-20z} dz\right]\) Let's evaluate these integrals: \(Q = 5\rho \int_{0}^{2\pi} d\phi \left[ \frac{1}{20}(e^{20z}\big|_{-\infty}^{0} - e^{-20z}\big|_{0}^{\infty})\right]\) \(Q = 5\rho \int_{0}^{2\pi} d\phi \left[\frac{1}{20}(1 + 1)\right]\) Finally, we can evaluate the integral: \(Q = \frac{1}{2} \rho \int_{0}^{2\pi} 5 d\phi\) \(Q = \frac{1}{2} \rho (10\pi) = 40\pi \mathrm{nC}\) So, the total amount of charge present on the cylindrical surface is \(40\pi\) nC.

Calculate Electric Flux

Next, we need to calculate the electric flux leaving the surface within the specified limits for z and φ. Electric flux is given by: \(Φ_E = \int\int \mathbf{E} \cdot d\mathbf{A}\) \(\rho_S = 5 e^{-20|z|}\) nC/m² is the surface charge density and the electric field is proportional to the surface charge density, so we can write: \(\mathbf{E} \cdot d\mathbf{A} = k \rho_S dA\), where k is a constant. Now we need to integrate this expression within the given limits: \(Φ_E = \rho k \int_{1 \mathrm{cm}}^{5\mathrm{cm}} dz \int_{30^\circ}^{90^\circ} 5 e^{-20|z|} d\phi\) The integral is separable, and we can calculate it as: \(Φ_E = 5\rho k \int_{1 \mathrm{cm}}^{5\mathrm{cm}} e^{-20z} dz \int_{30^\circ}^{90^\circ} d\phi\) Solving the integrals, we get: \(Φ_E = 5\rho k \left[ \frac{1}{20} (e^{-20(5\,\mathrm{cm}-1\, \mathrm{cm})} - e^{-20(1\,\mathrm{cm})})\right]\left[(90^\circ - 30^\circ)\right]\) \(Φ_E = 5\rho k \left[ \frac{1}{20} (e^{-80} - e^{-20}) \right] (60^\circ)\) To compute the final answer, we plug in the value for \(\rho\): MessageLookup

Key concepts

Surface Charge Density

Surface charge density is a critical concept in electromagnetism that deals with the amount of charge per unit area on a surface. In our cylindrical surface problem, the surface charge density is represented by the variable \(\rho_S\) and is given by the expression \(\rho_S = 5 e^{-20|z|} \mathrm{nC/m^2}\). This depicts how charge density varies with the distance along the z-axis in an exponential manner, decreasing as you move away from the center. When calculating the total charge present, we integrate this density over the entire surface area to account for every portion of charge across the surface. This integral is essential for understanding how much charge is distributed over the cylindrical surface and provides the groundwork for calculating electric properties like potential and field strength around the surface.

To enhance understanding, consider surface charge density similar to the coating of paint on a surface – where the density would indicate how thickly the paint is laid out over each unit area. In our case, it's the 'paint' of electric charge across our cylindrical surface.

Electric Flux

Electric flux is a measure of how much electric field passes through a given area. It's an important concept which helps understand how fields interact with surfaces. The mathematical representation of electric flux \(\Phi_E\) is integral to our problem, where we need to find the quantity of electric flux leaving a specified part of the charge-carrying surface.

In the context of our exercise, electric flux is proportional to the surface charge density. By integrating the surface charge density over a certain part of the surface area, using the formula \(\Phi_E = \int\int \mathbf{E} \cdot d\mathbf{A}\), we calculate the electric flux emanating from the surface within given z and φ limits. The concept can be visualized like the flow of water through a net; the water represents the electric field, while the net embodies the specific surface area through which we assess the flow.

Cylindrical Coordinates

Cylindrical coordinates are a means of describing the position of a point in a 3D space, particularly suited for objects with cylindrical symmetry, much like the surface in our problem. The system uses three values: the radial distance from the origin \(\rho\), the azimuthal angle \(\phi\) around the z-axis, and the height \(z\) along the z-axis.

For the problem at hand, we don't vary the radial distance \(\rho\) as it's given as 8 cm, but we do consider the variation in \(z\) along the height and \(\phi\) around the axis to find out the charge and flux within certain sections of the cylindrical surface. By using cylindrical coordinates, we inherently tailor the calculation processes such as integration, to the shape and symmetry of the object we're examining, thus simplifying our computations. It's akin to choosing the right tool for the job, ensuring efficiency and precision in our calculations.