Charged Spherical Shell

Claimed by Erin M. Claimed by Noah Sutter (Fall 2016)

The Main Idea

A charged spherical shell is referring to the idea that the charge on a sphere is spread equally over its surface in a thin layer that resembles a shell. Charged objects create electric fields and this electric field depends on its shape, charge, and the distance to the observation location. In the case of a charged spherical shell if the observation location is within shell (distance less than the radius of the sphere) the electric field is zero. If the observation location is outside of the shell (distance greater than the radius of the sphere) then the equation to calculate the electric field of a point charge can be applied.

A Mathematical Model

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the sphere: [math]\displaystyle{ \vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r} }[/math]

Observation location inside of the sphere: [math]\displaystyle{ \vec{E}_{sphere}=0 }[/math]

A Computational Model

Observation Location Outside Spherical Shell:

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this trinket for an example of an electric field of a spherical shell with observation locations outside of the shell.

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through Gauss's Law.

Gauss's Law states [math]\displaystyle{ \frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA }[/math].

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,

then the area can be replaced as follows:

[math]\displaystyle{ E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2} }[/math].

In other words, [math]\displaystyle{ E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2} }[/math] which is the equation for an electric field of a charged particle.

Observation Location Inside the Hollow Portion of the Spherical Shell:

If the observation location is anywhere inside of the spherical shell, then the electric field is zero.

insert figure

This figure shows a cross section of a charged spherical shell. As you can see there are positive charges distributed equally around the circle that contribute to the net electric field. Suppose we choose a point inside the sphere, point A.

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The contributions to the electric field at this point are shown below, considering eight equally spaced charges around the surface of the sphere.

insert figure

As you can see, the closer a charge is to point A inside the sphere, the smaller its contribution to the electric field at point A. This is because the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. However, because the surface area of the sphere is proportional to r^2 (surface area of sphere = 4pir^2), and because the charges are uniform, all the contributions cancel each other out. (1/r^2 * r^2 = 1). This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through Gauss's Law.

[math]\displaystyle{ \frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA }[/math]

Using a gaussian surface of a sphere of radius r where r<a<b

the charge enclosed by the gaussian surface is 0, [math]\displaystyle{ 0 = \oint_C E\bullet dA }[/math] and thus the electric field through the surface is also 0.

IMPORTANT: It is very important to remember that the electric field inside a sphere is NOT always zero. Other charges that exist may contribute to the electric field inside the sphere, only the electric field that is being caused by the uniformly distributed charges on the surface of the sphere is zero.

Observation Location Within the Spherical Shell:

The electric field at an observation location within the spherical shell itself is dependent on the percentage of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell. If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: [math]\displaystyle{ Q_1 = \frac{r^3-a^3}{b^3-a^3}Q }[/math]

Using Gauss's Law the electric field through the gaussian surface (a sphere of radius r):

[math]\displaystyle{ E = \frac{Q_1}{\epsilon_0*4\pi r^2} }[/math]

There is no electric field contributed by portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

Examples

Simple

A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

E=0

Middling

A spherical shell of charge with a radius of 5m is located at the origin and is uniformly charged with q=+2e-7. What is the electric field produced from the spherical shell at x=10m?

E_sphere = (1/4πε_0)(q/r^2)rhat

E_sphere = (1/4πε_0)(2e-7/5^2)<1,0,0>

E_sphere = <72,0,0>

Difficult

A spherical shell of charge with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

E_sphere = (1/4πε_0)(q/r^2)rhat

E_sphere = (1/4πε_0)(6e-8/7^2)<6/7,3/7,2/7>

E_sphere ≈ (11.02)<6/7,3/7,2/7>

E_sphere ≈ <9.45,4.72,3.15>

Connectedness

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

History

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

See also

Electric Force One application of electric fields due to point charges deals with finding electric force

Electric Field More general ideas about electric fields

Further reading

Principles of Electrodynamics by Melvin Schwartz ISBN: 9780486134673

External links

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

References

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