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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/(4蟺蔚鈧�r虏), where Q is the total charge on the shell, 蔚鈧� 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 (蔚鈧�). 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 4蟺r虏 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 4蟺r虏: \[ E = \frac{Q}{4\pi \epsilon_0 r^2}. \] The electric field outside the thin spherical shell is given by the expression E = Q/(4蟺蔚鈧�r虏).

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Most popular questions from this chapter

In fair weather, over flat ground, there is a downward electric field of about \(150 \mathrm{N} / \mathrm{C} .\) (a) Assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is $150 \mathrm{N} / \mathrm{C}$ pointing radially inward, calculate the total charge on the Earth and the charge per unit area. (b) At an altitude of $250 \mathrm{m}\( above Earth's surface, the field is only \)120 \mathrm{N} / \mathrm{C}$ Calculate the charge density (charge per unit volume) of the air (assumed constant). [Hint: See Conceptual Example \(16.8 .]\)
A small sphere with a charge of \(-0.60 \mu \mathrm{C}\) is placed in a uniform electric field of magnitude \(1.2 \times 10^{6} \mathrm{N} / \mathrm{C}\) pointing to the west. What is the magnitude and direction of the force on the sphere due to the electric field?
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.
(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 balloon, initially neutral, is rubbed with fur until it acquires a net charge of \(-0.60 \mathrm{nC} .\) (a) Assuming that only electrons are transferred, were electrons removed from the balloon or added to it? (b) How many electrons were transferred?
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