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Chapter 22: Problem 61

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

Expert verified
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

01

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.
02

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. \(\implies E = 0\;\mathrm{N/C}\)
03

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.
04

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.

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Most popular questions from this chapter

How is it possible that the flux through a closed surface does not depend on where inside the surface the charge is located (that is, the charge can be moved around inside the surface with no effect whatsoever on the flux)? If the charge is moved from just inside to just outside the surface, the flux changes discontinuously to zero, according to Gauss's Law. Does this really happen? Explain.

A hollow conducting spherical shell has an inner radius of \(8.00 \mathrm{~cm}\) and an outer radius of \(10.0 \mathrm{~cm} .\) The electric field at the inner surface of the shell, \(E_{\mathrm{i}}\), has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points toward the center of the sphere, and the electric field at the outer surface, \(E_{\infty}\) has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points away from the center of the sphere (see the figure). Determine the magnitude of the charge on the inner surface and the outer surface of the spherical shell.

An electric dipole consists of two equal and opposite charges situated a very small distance from each other. When the dipole is placed in a uniform electric field, which of the following statements is true? a) The dipole will not experience any net force from the electric field; since the charges are equal and have opposite signs, the individual effects will cancel out. b) There will be no net force and no net torque acting on the dipole. c) There will be a net force but no net torque acting on the dipole. d) There will be no net force, but there will (in general) be a net torque acting on dipole.

A dipole is completely enclosed by a spherical surface. Describe how the total electric flux through this surface varies with the strength of the dipole.

A cube has an edge length of \(1.00 \mathrm{~m} .\) An electric field acting on the cube from outside has a constant magnitude of \(150 \mathrm{~N} / \mathrm{C}\) and its direction is also constant but unspecified (not necessarily along any edges of the cube). What is the total charge within the cube?
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