Question

A sphere centered at the origin has a volume charge distribution of \(120 \mathrm{nC} / \mathrm{cm}^{3}\) and a radius of \(12 \mathrm{~cm}\). The sphere is centered inside a conducting spherical shell with an inner radius of \(30.0 \mathrm{~cm}\) and an outer radius of \(50.0 \mathrm{~cm}\). The charge on the spherical shell is \(-2.0 \mathrm{mC}\). What is the magnitude and direction of the electric field at each of the following distances from the origin? a) at \(r=10.0 \mathrm{~cm}\) c) at \(r=40.0 \mathrm{~cm}\) b) at \(r=20.0 \mathrm{~cm}\) d) at \(r=80.0 \mathrm{~cm}\)

Short answer

Question: Calculate the magnitude and direction of the electric field at various distances (r = 10 cm, 20 cm, 40 cm, and 80 cm) from the origin due to a charged sphere with charge density of \(120 \;\text{nC/cm}^3\) and radius of 12 cm, and a conducting spherical shell with inner and outer radii of 30 cm and 50 cm, respectively, with a total charge of \(-2\) mC. Answer: At r = 10 cm, the electric field's magnitude is \(\frac{120\times10^{-9}}{3\epsilon_0} \;\text{N/C}\), with a direction radially outward. At r = 20 cm, the electric field's magnitude is 0 N/C. At r = 40 cm, the electric field's magnitude is \(\frac{\frac{4 \times 120\times10^{-9}\times\pi\times(0.12)^3}{3} - 2\times10^{-3}}{4\pi\epsilon_0 (0.4)^2}\;\text{N/C}\), with a direction radially inward. At r = 80 cm, the electric field's magnitude is \(\frac{\frac{4 \times 120\times10^{-9}\times\pi\times(0.12)^3}{3} - 2\times10^{-3}}{4\pi\epsilon_0 (0.8)^2} \;\text{N/C}\), with a direction radially inward.

Step-by-step solution

a) At r = 10 cm

The distance \(r = 10\;\mathrm{cm} = 0.1\;\mathrm{m}\). Since this distance is within the charged sphere, the enclosed charge can be calculated using the volume charge density. Let's find the charge enclosed up to 0.1 m: \(\implies Q_{\text{enc}} = \rho V = 120\times10^{-9} \frac{4\pi}{3}(0.1)^3 = \frac{4 \times 120 \times 10^{-9} \times \pi \times (0.1)^3}{3}\) Now, we need to find the electric field using Gauss's law. We consider a Gaussian surface as a sphere with radius of \(r = 0.1\) m. \(\implies \oint \vec{E}\cdot \vec{dA} = E \oint dA = E 4\pi (0.1)^2 = \frac{Q_{\text{enc}}}{\epsilon_0}\) \(\implies E = \frac{Q_{\text{enc}}}{4\pi \epsilon_0 (0.1)^2}\) Putting the values and calculating E: \(E = \frac{4 \times 120 \times 10^{-9} \times \pi \times (0.1)^3/(3)}{4\pi \epsilon_0 (0.1)^2} = \frac{120\times 10^{-9}}{3\epsilon_0} \;\mathrm{N/C}\) The direction of the electric field is radially outside, as the enclosed charge is positive.

b) At r = 20 cm

The distance \(r = 20\;\mathrm{cm} = 0.2\;\mathrm{m}\). Since this distance falls inside the spherical shell, the electric field within the conducting shell is always zero. \(\implies E = 0\;\mathrm{N/C}\)

c) At r = 40 cm

The distance \(r = 40\;\mathrm{cm} = 0.4\;\mathrm{m}\). At this distance, the point is between the two spheres and the enclosed charge includes both the sphere and the conducting shell. So, the total enclosed charge is the sum of charges of the volume charge distribution and the charge on the spherical shell. \(Q_{\text{enc}} = \frac{4 \times 120\times 10^{-9} \times \pi \times (0.12)^3}{3} + (-2\times10^{-3})\) Now, we consider a Gaussian surface as a sphere with a radius of \(r=0.4\) m. \(\implies \oint \vec{E}\cdot \vec{dA} = E\oint dA = E 4\pi (0.4)^2 = \frac{Q_{\text{enc}}}{\epsilon_0}\) \(\implies E = \frac{Q_{\text{enc}}}{4\pi \epsilon_0 (0.4)^2}\) Putting the values and calculating E: \(E = \frac{\frac{4 \times 120 \times 10^{-9} \times \pi \times (0.12)^3}{3} - 2\times10^{-3}}{4\pi\epsilon_0 (0.4)^2} \;\mathrm{N/C}\) The direction of the electric field is radially inward, as the net enclosed charge is negative.

d) At r = 80 cm

At r = 80 cm \(=0.8\;\mathrm{m}\), we need to find the electric field for a point outside both the sphere and the shell. Since the total enclosed charge is equal to the sum of charges, we can use the same enclosed charge as in point c. We consider a Gaussian surface as a sphere with a radius of \(r=0.8\;\mathrm{m}\). \(\implies \oint \vec{E}\cdot \vec{dA} = E \oint dA = E 4\pi (0.8)^2 = \frac{Q_{\text{enc}}}{\epsilon_0}\) \(\implies E = \frac{Q_{\text{enc}}}{4\pi \epsilon_0 (0.8)^2}\) Putting the values and calculating E: \(E = \frac{\frac{4 \times 120 \times 10^{-9} \times \pi \times (0.12)^3}{3} - 2\times10^{-3}}{4\pi\epsilon_0 (0.8)^2} \;\mathrm{N/C}\) The direction of the electric field is radially inward, as the net enclosed charge is negative.

Key concepts

Gauss's Law

Gauss's law is a fundamental principle in electromagnetism that relates the electric field flowing out of a closed surface to the charge enclosed by that surface. It can be represented mathematically as: \[ \oint \vec{E} \cdot \vec{dA} = \frac{Q_{\text{enc}}}{\epsilon_0} \] where

• \( \vec{E} \) is the electric field vector.
• \( \vec{dA} \) is the differential area vector on the closed surface.
• \( Q_{\text{enc}} \) is the total charge enclosed by the surface.
• \( \epsilon_0 \) is the permittivity of free space.

This law is particularly useful for calculating electric fields when dealing with symmetrical charge distributions, such as spheres or cylinders. To use Gauss's law, choose a Gaussian surface that aligns with the symmetry of the problem. This simplifies calculations as it allows the electric field \( \vec{E} \) to be constant over the surface or zero inside conductors.

Volume Charge Density

Volume charge density describes how much electric charge is present per unit volume in a region of space. It is usually denoted as \( \rho \) and is expressed in units of coulombs per cubic meter (C/m³).
In the context of a sphere, the total charge \( Q \) inside can be calculated by multiplying the charge density by the volume of the sphere:\[ Q = \rho \times V \] For a sphere with radius \( r \), the volume \( V \) is \( \frac{4}{3}\pi r^3 \). Therefore, the total enclosed charge is \[ Q = \rho \left( \frac{4}{3} \pi r^3 \right) \]This measurement is crucial in determining how the electric field behaves inside a material with gradient or uniform charge distribution. For example, if the volume charge density is high, it implies that a large amount of charge is packed in a smaller space, leading to a stronger electric field.

Conducting Spherical Shell

A conducting spherical shell is composed of a conductive material and has an inner and an outer radius. When charges are placed on the shell, they redistribute themselves on its outer surface due to the conductive property of the material.
Key properties include:

• The electric field inside the shell, within the hollow region, is zero. This happens because any excess charge resides entirely on the outer surface, keeping the interior neutral thanks to the electrostatic shielding.
• If there is any charge inside the shell that is not on the surface, it does not affect the field outside. Instead, the outer field depends only on the total charge on the outer surface.

These features make conducting spherical shells useful in applications like Faraday cages, which shield sensitive electronic equipment from external electric fields.

Enclosed Charge

Enclosed charge refers to the total charge contained within a given closed surface. This is a pivotal concept in Gauss's law, where the total enclosed charge affects the resulting electric field.
To find the enclosed charge:- Consider all the charge contributions within the Gaussian surface.- Calculate using volume charge density and any additional contributions from other parts, such as from adjacent shell materials or structures.
For example, within a concentric system of a charged sphere and a shell, the enclosed charge could be the sum of the charge inside the sphere and any induced charges on the shell. Thus, calculating \( Q_{\text{enc}} \) requires evaluating contributions from both the volume charge and surface charges, such as the specific charge on the shell, which must be accounted for. This comprehensive approach ensures that the influence on the electric field is accurately determined.