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Chapter 21: Problem 52

A thick, spherical shell of inner radius \(a\) and outer radius \(b\) carries a uniform volume charge density \(\rho .\) Find an expression for the electric field strength in the region \(a

Short Answer

Expert verified
The electric field strength in the region \(a<r<b\) is \(E=\frac{\rho}{3\epsilon_{0}} \cdot (r - \frac{a^3}{r^2})\) and it's consistent with Equation 21.5 when \(a=0\), which provides \(E=\frac{\rho}{3\epsilon_{0}} \cdot r \) for electric field within a uniformly charged sphere.

Step by step solution

01

Set up Integral for Gauss's Law

To find the electric field, we will need to integrate across the volume of the spherical shell using Gauss's Law, which is \(\oint \textbf{E} \cdot d\textbf{a} = \frac{Q_{\text{enc}}}{\epsilon_{0}}\). Here, \(\textbf{E}\) is the electric field, \(d\textbf{a}\) is a small area vector on the gaussian surface, \(Q_{\text{enc}}\) is the charge enclosed by the gaussian surface, and \(\epsilon_{0}\) is the permittivity of free space. Our Gaussian surface, in this case, is a sphere of radius \(r\) where \(a<r<b\). Notice that the electric field \(\textbf{E}\) and the area vector \(d\textbf{a}\) are in the same direction for a sphere.
02

Enclosed Charge Calculation

The enclosed charge within the Gaussian surface is given by the volume of that surface times the charge density, or \(Q_{\text{enc}} = \rho \cdot V_{\text{enc}} = \rho \cdot \frac{4}{3} \pi r^3 - \rho \cdot \frac{4}{3} \pi a^3\). Here, we are subtracting the volume of the inner sphere of radius \(a\) from the volume of the Gaussian surface of radius \(r\). The charge is homogeneously distributed, so the charge density \(\rho\) is the same throughout.
03

Electric Field Calculation

Substitute the value of \(Q_{\text{enc}}\) into Gauss's Law: \(\oint \textbf{E} \cdot d\textbf{a} = \frac{Q_{\text{enc}}}{\epsilon_{0}} \\ \Rightarrow E \cdot 4\pi r^2 = \frac{\rho}{\epsilon_{0}} \cdot (\frac{4}{3} \pi r^{3} - \frac{4}{3} \pi a^{3})\\ \Rightarrow E=\frac{\rho}{3\epsilon_{0}} \cdot (r - \frac{a^3}{r^2}) \). This is the electric field strength in the region \(a<r<b\).
04

Verifying with Equation 21.5

When \(a=0\), this implies there's no internal hollow space. Substituting this into our derived equation, we get: \(E=\frac{\rho}{3\epsilon_{0}} \cdot r \), which is consistent with Equation 21.5 (the electric field inside a uniformly charged sphere).

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Most popular questions from this chapter

A rod \(50 \mathrm{cm}\) long and \(1.0 \mathrm{cm}\) in radius carries a \(2.0-\mu \mathrm{C}\) charge distributed uniformly over its length. Find the approximate magnitude of the electric field (a) \(4.0 \mathrm{mm}\) from the rod surface, not near either end, and (b) 23 m from the rod.

A \(2.6-\mu \mathrm{C}\) charge is at the center of a cube \(7.5 \mathrm{cm}\) on each side. What's the electric flux through one face of the cube? (Hint: Think about symmetry, and don't do an integral.)

A flat surface with area \(2.0 \mathrm{m}^{2}\) is in a uniform \(850-\mathrm{N} / \mathrm{C}\) electric field. Find the electric flux through the surface when it's (a) at right angles to the field, (b) at \(45^{\circ}\) to the field, and (c) parallel to the field.

The electric field in a certain region is given by \(\vec{E}=a x \hat{\imath},\) where \(a=40 \mathrm{N} / \mathrm{C} \cdot \mathrm{m}\) and \(x\) is in meters. Find the volume charge density in the region. (Hint: Apply Gauss's law to a cube 1 m on a side.)

What's the electric field strength in a region where the flux through a \(1.0 \mathrm{cm} \times 1.0 \mathrm{cm}\) flat surface is \(65 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) if the field is uniform and the surface is at right angles to the field?
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