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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

A total charge of \(7.50 \times 10^{-6} \mathrm{C}\) is distributed on two different small metal spheres. When the spheres are \(6.00 \mathrm{cm}\) apart, they each feel a repulsive force of \(20.0 \mathrm{N} .\) How much charge is on each sphere?
A parallel-plate capacitor consists of two flat metal plates of area \(A\) separated by a small distance \(d\). The plates are given equal and opposite net charges \(\pm q\) (a) Sketch the field lines and use your sketch to explain why almost all of the charge is on the inner surfaces of the plates. (b) Use Gauss's law to show that the electric field between the plates and away from the edges is $E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} \cdot(\mathrm{c})$ Does this agree with or contra- dict the result of Problem \(70 ?\) Explain. (d) Use the principle of superposition and the result of Problem 69 to arrive at this same answer. [Hint: The inner surfaces of the two plates are thin, flat sheets of charge.]
A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$
The electric field across a cellular membrane is $1.0 \times 10^{7} \mathrm{N} / \mathrm{C}$ directed into the cell. (a) If a pore opens, which way do sodium ions (Na") flow- into the cell or out of the cell? (b) What is the magnitude of the electric force on the sodium ion? The charge on the sodium ion is \(+e\)
What is the ratio of the electric force to the gravitational force between a proton and an electron separated by \(5.3 \times 10^{-11} \mathrm{~m}\) (the radius of a hydrogen atom)?
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