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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

A thin, hollow, metal cylinder of radius \(R\) has a surface charge distribution \(\sigma\). A long, thin wire with a linear charge density \(\lambda / 2\) runs through the center of the cylinder. Find an expression for the electric fields and the direction of the field at each of the following locations: a) \(r \leq R\) b) \(r \geq R\)

An object with mass \(m=1.0 \mathrm{~g}\) and charge \(q\) is placed at point \(A\), which is \(0.05 \mathrm{~m}\) above an infinitely large, uniformly charged, nonconducting sheet \(\left(\sigma=-3.5 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}\right)\), as shown in the figure. Gravity is acting downward \(\left(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\). Determine the number, \(N\), of electrons that must be added to or removed from the object for the object to remain motionless above the charged plane.

Consider an electric dipole on the \(x\) -axis and centered at the origin. At a distance \(h\) along the positive \(x\) -axis, the magnitude of electric field due to the electric dipole is given by \(k(2 q d) / h^{3} .\) Find a distance perpendicular to the \(x\) axis and measured from the origin at which the magnitude of the electric field stays the same.

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 point charge, \(+Q\), is located on the \(x\) -axis at \(x=a\), and a second point charge, \(-Q\), is located on the \(x\) -axis at \(x=-a\). A Gaussian surface with radius \(r=2 a\) is centered at the origin. The flux through this Gaussian surface is a) zero. b) greater than zero. c) less than zero. d) none of the above.
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