Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?
Point charges are arranged on the vertices of a square with sides of $2.50 \mathrm{cm} .$ Starting at the upper left corner and going clockwise, we have charge A with a charge of \(0.200 \mu \mathrm{C}, \mathrm{B}\) with a charge of \(-0.150 \mu \mathrm{C}, \mathrm{C}\) with a charge of \(0.300 \mu \mathrm{C},\) and \(\mathrm{D}\) with a mass of \(2.00 \mathrm{g},\) but with an unknown charge. Charges \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are fixed in place, and \(\mathrm{D}\) is free to move. Particle D's instantaneous acceleration at point \(D\) is \(248 \mathrm{m} / \mathrm{s}^{2}\) in a direction \(30.8^{\circ}\) below the negative \(x\) -axis. What is the charge on D?
\(\mathrm{A}+2.0\) -nC point charge is \(3.0 \mathrm{cm}\) away from a $-3.0 \mathrm{-nC}$ point charge. (a) What are the magnitude and direction of the electric force acting on the +2.0 -nC charge? (b) What are the magnitude and direction of the electric force acting on the -3.0 -nC charge?
Two Styrofoam balls with the same mass \(m=9.0 \times 10^{-8} \mathrm{kg}\) and the same positive charge \(Q\) are suspended from the same point by insulating threads of length \(L=0.98 \mathrm{m} .\) The separation of the balls is \(d=0.020 \mathrm{m} .\) What is the charge \(Q ?\)

A metal sphere A has charge \(Q\). Two other spheres, B and \(\mathrm{C},\) are identical to \(\mathrm{A}\) except they have zero net charge. A touches \(\mathrm{B}\), then the two spheres are separated. B touches \(C\), then those spheres are separated. Finally, C touches A and those two spheres are separated. How much charge is on each sphere?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free