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Chapter 16: Problem 73

Use Gauss's law to derive an expression for the electric field outside the thin spherical shell of Conceptual Example 16.8

Short Answer

Expert verified
Short Answer: The electric field (E) outside a thin spherical shell is given by the expression E = Q/(4πε₀r²), where Q is the total charge on the shell, ε₀ is the permittivity of free space, and r is the distance from the center of the shell. This is derived using Gauss's Law and considering a Gaussian surface outside the shell.

Step by step solution

01

Understand Gauss's Law

Gauss's law states that the electric flux through a closed surface is equal to the total enclosed charge divided by the permittivity of free space (ε₀). Mathematically, \[ \oint \vec{E}\cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}. \]
02

Set up Gaussian Surface

Given a thin spherical shell, we want to find the electric field outside the shell. To do this, consider a larger sphere centered on the shell, which will be the Gaussian surface. The radius of this Gaussian surface will be r, where r is greater than the radius of the shell, R.
03

Calculate the Electric Flux

The electric field, E, is radial; thus, it can only have a component along the area vector, dA. This simplifies the dot product: \[ \oint \vec{E}\cdot d\vec{A} = \oint EdA. \] Since the electric field is uniform over the Gaussian surface (the sphere), we can move E out of the integral: \[ E \oint dA = E (4\pi r^2), \] where 4πr² is the surface area of the Gaussian sphere.
04

Determine the Enclosed Charge

The total charge enclosed by the Gaussian sphere is the same as the total charge on the thin spherical shell, Q, since we are considering a point outside the shell.
05

Apply Gauss's Law

Now we can use Gauss's law to relate the electric field to the enclosed charge. Substitute the electric flux and enclosed charge expressions into Gauss's law: \[ E (4\pi r^2) = \frac{Q}{\epsilon_0}. \]
06

Solve for the Electric Field

Finally, solve for the electric field E by dividing both sides of the equation by 4πr²: \[ E = \frac{Q}{4\pi \epsilon_0 r^2}. \] The electric field outside the thin spherical shell is given by the expression E = Q/(4πε₀r²).

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Most popular questions from this chapter

The Bohr model of the hydrogen atom proposes that the electron orbits around the proton in a circle of radius \(5.3 \times 10^{-11} \mathrm{m} .\) The electric force is responsible for the radial acceleration of the electron. What is the speed of the electron in this model?
What are the magnitude and direction of the acceleration of a proton at a point where the electric field has magnitude \(33 \mathrm{kN} / \mathrm{C}\) and is directed straight up?
(a) Find the electric flux through each side of a cube of edge length \(a\) in a uniform electric field of magnitude \(E\) The field direction is perpendicular to two of the faces. (b) What is the total flux through the cube?
A dipole consists of two equal and opposite point charges \((\pm q)\) separated by a distance \(d .\) (a) Write an expression for the electric field at a point \((0, y)\) on the dipole axis. Specify the direction of the field in all four regions: \(y>\frac{1}{2} d, 0<y<\frac{1}{2} d,-\frac{1}{2} d<y<0,\) and \(y<-\frac{1}{2} d .\) (b) At distant points \((|y| \gg d),\) write a simpler, approximate expression for the field. To what power of \(y\) is the field proportional? Does this conflict with Coulomb's law? [Hint: Use the binomial approximation $\left.(1 \pm x)^{n} \approx 1 \pm n x \text { for } x \ll 1 .\right]$
Two tiny objects with equal charges of \(7.00 \mu \mathrm{C}\) are placed at the two lower corners of a square with sides of \(0.300 \mathrm{m},\) as shown. Find the electric field at point \(C\) the center of the square.
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