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Chapter 22: Problem 84

A uniform sphere has a radius \(R\) and a total charge \(+Q,\) uniformly distributed throughout its volume. It is surrounded by a thick spherical shell carrying a total charge \(-Q,\) also uniformly distributed, and having an outer radius of \(2 R\). What is the electric field as a function of \(R ?\)

Short Answer

Expert verified
Question: Determine the electric field as a function of the radius R for a uniformly charged sphere with total charge +Q and radius R, surrounded by a uniformly charged spherical shell with outer radius 2R and total charge -Q. Answer: The electric field as a function of R is given by: $$ E(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$

Step by step solution

01

Determine the electric field of a uniformly charged sphere

First, let's find the electric field of a uniformly charged sphere. To do this, we use Gauss' Law: $$ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} $$ For a sphere of radius r (with r <= R) centered at the origin, the enclosed charge is $$ Q_{inclosed} = \frac{Q}{\frac{4}{3}\pi R^3} \times \frac{4}{3}\pi r^3 $$ Then, the electric field is same for each point on the spherical surface enclosing the uniformly charged sphere, so: $$ E \times 4\pi r^2 = \frac{Q}{\frac{4}{3}\pi R^3} \times \frac{4}{3}\pi r^3 \div \epsilon_0 $$ Solving for E, we get: $$ E_{sphere}(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$
02

Determine the electric field of a uniformly charged spherical shell

Now, let's find the electric field of a uniformly charged spherical shell. To do this, we again use Gauss' Law, but this time we'll use a Gaussian surface that is a sphere centered at the origin, with 2R >= r >= R. Since there is no charge enclosed by this Gaussian surface, we have: $$ \oint \vec{E} \cdot d\vec{A} = 0 $$ Then, we can say that the electric field due to the shell at the Gaussian surface is canceled by the electric field due to the inner uniformly charged sphere: $$ E_{shell}(r) = -E_{sphere}(r) $$
03

Combine the electric fields

Now, we combine the electric fields from step 1 and step 2 to obtain the total electric field as a function of R: For r <= R, the electric field is only due to the charged sphere, so: $$ E(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$ For R <= r <= 2R, the electric field is due to both the sphere and the shell, so, as we derived, the electric field is: $$ E(r) = -E_{shell}(r) = -\left(-\frac{Qr}{4\pi\epsilon_0 R^3}\right) = \frac{Qr}{4\pi\epsilon_0 R^3} $$ Thus, the electric field as a function of R is: $$ E(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$

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Most popular questions from this chapter

A thin glass rod is bent into a semicircle of radius \(R\). A charge \(+Q\) is uniformly distributed along the upper half, and a charge \(-Q\) is uniformly distributed along the lower half as shown in the figure. Find the magnitude and direction of the electric field \(\vec{E}\) (in component form) at point \(P\), the center of the semicircle.

An electron is observed traveling at a speed of \(27.5 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) parallel to an electric field of magnitude \(11,400 \mathrm{~N} / \mathrm{C}\) How far will the electron travel before coming to a stop?

Electric dipole moments of molecules are often measured in debyes \((\mathrm{D}),\) where \(1 \mathrm{D}=3.34 \cdot 10^{-30} \mathrm{C} \mathrm{m} .\) For instance, the dipole moment of hydrogen chloride gas molecules is \(1.05 \mathrm{D}\). Calculate the maximum torque such a molecule can experience in the presence of an electric field of magnitude \(160.0 \mathrm{~N} / \mathrm{C}\).

Consider a long horizontally oriented conducting wire with \(\lambda=4.81 \cdot 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton \(\left(\mathrm{mass}=1.67 \cdot 10^{-27} \mathrm{~kg}\right)\) is placed \(0.620 \mathrm{~m}\) above the wire and released. What is the magnitude of the initial acceleration of the proton?

A charge of \(+2 q\) is placed at the center of an uncharged conducting shell. What will be the charges on the inner and outer surfaces of the shell, respectively? a) \(-2 q,+2 q\) b) \(-q,+q\) c) \(-2 q,-2 q\) d) \(-2 q,+4 q\)
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