Problem 25 - University Physics with Modern Physics

Question

22.25. The electric field at a distance of 0.145 \(\mathrm{m}\) from the surface of a solid insulating sphere with radius 0.355 \(\mathrm{m}\) is 1750 \(\mathrm{N} / \mathrm{C}\) . (a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? ( 6 ) Calculate the electric field inside the sphere at a distance of 0.200 \(\mathrm{m}\) from the center.

Step-by-Step Solution

Answer

Charge density is approximately \(2.59 \times 10^{-7} \ \mathrm{C}/\mathrm{m}^3\), the electric field inside at 0.200 m is approximately 1956 N/C.

Step 1: Understand the concepts

To solve this, we need to understand that the charge is uniformly distributed inside a solid insulating sphere. The problem involves finding charge density and the electric field using Gauss's Law for a charge distribution.

Step 2: Calculate total charge using Gauss's Law and known field

Given that the electric field at a distance of 0.500 m (0.145 m from the surface) is 1750 N/C. By Gauss's Law, the electric field outside the sphere is given by:\[E = \frac{kQ}{r^2}\]where - \(E\) is the electric field 1750 N/C,- \(k\) is the Coulomb's constant, \( 8.99 \times 10^9 \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2 \),- \(r = 0.355 + 0.145 = 0.500\mathrm{m}\) is the total distance from the center.Thus, solve for total charge \(Q\):\[Q = \frac{E \cdot r^2}{k} = \frac{1750 \cdot (0.500)^2}{8.99 \times 10^9}\approx 4.86 \times 10^{-8} \ \mathrm{C}\]

Step 3: Calculate charge density

Charge density \(\rho\) is the total charge divided by the volume of the sphere \(V\). The volume is given by \[V = \frac{4}{3} \pi R^3\]where \(R=0.355\,\mathrm{m}\) is the radius of the sphere.Calculate \(V\):\[V = \frac{4}{3} \pi (0.355)^3 \approx 0.1878 \ \mathrm{m}^3\]Thus, the charge density \(\rho\) is:\[\rho = \frac{Q}{V} = \frac{4.86 \times 10^{-8}}{0.1878} \approx 2.59 \times 10^{-7} \ \mathrm{C}/\mathrm{m}^3\]

Step 4: Use Gauss's law to calculate electric field inside the sphere

When calculating the electric field inside the sphere at a distance of 0.200 m from the center, Gauss's Law states:\[E(inside) = \frac{kQ_{enc}}{r^2}\]where \(Q_{enc}\) is the charge enclosed, calculated from the charge density \(\rho\) and volume of the part of the sphere included (radius up to 0.200 m). Therefore, \(Q_{enc} = \rho \cdot V_{part}\).Calculate \(V_{part}\):\[V_{part} = \frac{4}{3} \pi (0.200)^3 \approx 0.0335\ \mathrm{m}^3\]Find \(Q_{enc}\):\[Q_{enc} = 2.59 \times 10^{-7} \times 0.0335 \approx 8.68 \times 10^{-9} \ \mathrm{C}\]Finally, calculate the electric field:\[E = \frac{8.99 \times 10^9 \times 8.68 \times 10^{-9}}{(0.200)^2} \approx 1956 \ \mathrm{N}/\mathrm{C}\]

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle that helps us understand how electric fields behave. It relates the electric flux passing through a closed surface to the charge enclosed by that surface. Imagine it as a magical tool that allows us to calculate the strength of electric fields without having to know every small detail about how the charges are distributed. The law is particularly convenient for problems with a high degree of symmetry, such as spheres, cylinders, and planes.The general formula is:\[ \Phi = \oint E \, \cdot \, dA = \frac{Q_{enc}}{\varepsilon_0} \]where:- \(\Phi\) is the electric flux,- \(E\) denotes the electric field,- \(dA\) is a small element of the area,- \(Q_{enc}\) indicates the enclosed charge, and- \(\varepsilon_0\) represents the permittivity of free space.For a sphere, like in our problem, it simplifies to calculating the electric field a certain distance from the center by only considering the charge enclosed within that imaginary spherical surface. This makes Gauss's Law highly effective when dealing with symmetrical shapes like an insulating sphere.

Charge Density

Charge density is a measure of how much charge is contained in a specific space. It tells you how compact the charge is within a given volume.In mathematical terms, the charge density \(\rho\) signifies the amount of charge per unit volume. It's calculated by dividing the total charge \(Q\) of an object by its volume \(V\):\[ \rho = \frac{Q}{V} \]When dealing with spherical objects, such as an insulating sphere, you can calculate the volume using the sphere's radius \(R\):\[ V = \frac{4}{3} \pi R^3 \]For uniform charge distribution inside a sphere, this formula helps to determine the charge density easily. Knowing the charge density helps us further analyze how an electric field is influenced inside and outside the sphere.It's crucial in problems like these since understanding the charge density allows us to find the enclosed charge for any subsection of the sphere, which is essential in applying Gauss's Law.

Insulating Sphere

An insulating sphere is a spherical object that doesn’t allow electric charges to move freely through its surface. It's different from a conducting sphere, where charges spread out evenly. In the context of this exercise, the sphere has a radius and contains a uniformly distributed charge. This uniform distribution simplifies calculations because the charge can be thought of as being "smeared out" evenly throughout the sphere. When analyzing electric fields with such a sphere, two zones are crucial: - **Inside the Sphere:** The electric field depends on the distance from the center of the sphere. Gauss's Law applies to find the field by considering the charge contained within a smaller hypothetical sphere inside the actual sphere. - **Outside the Sphere:** The entire charge of the sphere is considered in the calculation, as if it were concentrated at the center, mimicking a point charge. Understanding these principles enables us to calculate how the electric field behaves at any point inside or outside the spherical object.

Coulomb's Constant

Coulomb's Constant is a key factor in calculating the electric field created by a point charge. In the world of electric forces, it is a proportional constant that appears in Coulomb's Law.Coulomb's Law describes how the electric force between two distant charges decreases with the square of the distance between them. The formula is:\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where:- \(F\) is the force between the charges,- \(q_1\) and \(q_2\) are the charges,- \(r\) is the distance separating the charges, and- \(k = 8.99 \times 10^9 \, \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2\) is Coulomb's Constant.Coulomb's Constant ensures that the units of electric force calculations match up in the International System of Units (SI). It's integral in calculating the electric field around charged objects by providing a way to translate charge and distance into the field's strength.