Physics Book - User contributions [en]

Field of a Charged Ball

2015-12-03T03:01:16Z

Eerwood3:

By Eric Erwood

In this section, the electric field of a sphere charged throughout its volume will be discussed.

==The Main Idea==

In this section, we will focus on a scenario where a sphere has charge distributed throughout the entire object. Calculating the electric field both outside and inside the sphere will be addressed.

===A Mathematical Model===

Step 1: Given a solid charged sphere throughout its volume, the first step is to cut up the the sphere into pieces. As a result, the solid sphere will appear as a series of spherical shells.

Step 2: Relationship between r and R.
Next, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. All the spherical shells appear as point charges at the center of the sphere. As a result, the electric field outside the sphere is a point charge:

<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>
when r>R, and R is the radius of the sphere.

However, when r<R, the observation location is inside some of the shells but outside others.
To find <math>\vec E_{net} </math>, add the contributions to the electric field from the inner shells.
After adding the contributions of each inner shell, you should have an electric field equal to:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2}</math>

We find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

We find that:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

The charge inside the sphere is proportional to r. When r=R,

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} </math>

===A Computational Model===

https://trinket.io/glowscript/48f6efd07a

use this code to see how it looks, change the r value.

==Examples==

===Simple===
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.

Step 1: cut up the sphere into shells

step 2: we know that r<R

Next find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C </math>

===Difficult===
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)

[[File:wikiimage3f.jpg]]

For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud

(s « R in the diagram). (Use the following as necessary: R and ε0.)

Useful equations:

<math> \vec p = \alpha *E</math>

<math> \vec p = Q*s </math>

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find <math> \alpha </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s </math>

<math>\alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 </math>

==Looking Ahead, Gauss's Law==

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

<math>\oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} </math>

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

<math>\vec E * 4\pi*r^2= \frac{1}{\epsilon_0}Q\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} </math>

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

== See also ==

===Further reading===

http://www.physicsbook.gatech.edu/Point_Charge

http://www.physicsbook.gatech.edu/Electric_Dipole

http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem

==References==

This section contains the the references you used while writing this page

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

https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/

Matter and Interactions, 4th Edition: 1-2

Field of a Charged Ball

2015-12-03T02:59:36Z

Eerwood3: /* Difficult */

By Eric Erwood

In this section, the electric field due of sphere charged throughout its volume will be discussed.

==The Main Idea==

In this section, we will focus on a scenario where a sphere has charge distributed throughout the entire object. Calculating the electric field both outside and inside the sphere will be addressed.

===A Mathematical Model===

Step 1: Given a solid charged sphere throughout its volume, the first step is to cut up the the sphere into pieces. As a result, the solid sphere will appear as a series of spherical shells.

Step 2: Relationship between r and R.
Next, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. All the spherical shells appear as point charges at the center of the sphere. As a result, the electric field outside the sphere is a point charge:

<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>
when r>R, and R is the radius of the sphere.

However, when r<R, the observation location is inside some of the shells but outside others.
To find <math>\vec E_{net} </math>, add the contributions to the electric field from the inner shells.
After adding the contributions of each inner shell, you should have an electric field equal to:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2}</math>

We find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

We find that:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

The charge inside the sphere is proportional to r. When r=R,

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} </math>

===A Computational Model===

https://trinket.io/glowscript/48f6efd07a

use this code to see how it looks, change the r value.

==Examples==

===Simple===
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.

Step 1: cut up the sphere into shells

step 2: we know that r<R

Next find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C </math>

===Difficult===
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)

[[File:wikiimage3f.jpg]]

For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud

(s « R in the diagram). (Use the following as necessary: R and ε0.)

Useful equations:

<math> \vec p = \alpha *E</math>

<math> \vec p = Q*s </math>

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find <math> \alpha </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s </math>

<math>\alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 </math>

==Looking Ahead, Gauss's Law==

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

<math>\oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} </math>

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

<math>\vec E * 4\pi*r^2= \frac{1}{\epsilon_0}Q\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} </math>

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

== See also ==

===Further reading===

http://www.physicsbook.gatech.edu/Point_Charge

http://www.physicsbook.gatech.edu/Electric_Dipole

http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem

==References==

This section contains the the references you used while writing this page

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

https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/

Matter and Interactions, 4th Edition: 1-2

Field of a Charged Ball

2015-12-02T18:57:16Z

Eerwood3: /* References */

By Eric Erwood

In this section, the electric field due of sphere charged throughout its volume will be discussed.

==The Main Idea==

In this section, we will focus on a scenario where a sphere has charge distributed throughout the entire object. Calculating the electric field both outside and inside the sphere will be addressed.

===A Mathematical Model===

Step 1: Given a solid charged sphere throughout its volume, the first step is to cut up the the sphere into pieces. As a result, the solid sphere will appear as a series of spherical shells.

Step 2: Relationship between r and R.
Next, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. All the spherical shells appear as point charges at the center of the sphere. As a result, the electric field outside the sphere is a point charge:

<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>
when r>R, and R is the radius of the sphere.

However, when r<R, the observation location is inside some of the shells but outside others.
To find <math>\vec E_{net} </math>, add the contributions to the electric field from the inner shells.
After adding the contributions of each inner shell, you should have an electric field equal to:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2}</math>

We find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

We find that:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

The charge inside the sphere is proportional to r. When r=R,

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} </math>

===A Computational Model===

https://trinket.io/glowscript/48f6efd07a

use this code to see how it looks, change the r value.

==Examples==

===Simple===
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.

Step 1: cut up the sphere into shells

step 2: we know that r<R

Next find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C </math>

===Difficult===
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)
[[File:wikiimage3f.jpg]]

For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud

(s « R in the diagram). (Use the following as necessary: R and ε0.)

Useful equations:

<math> \vec p = \alpha *E</math>

<math> \vec p = Q*s </math>

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find <math> \alpha </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s </math>

<math>\alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 </math>

==Looking Ahead, Gauss's Law==

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

<math>\oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} </math>

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

<math>\vec E * 4\pi*r^2= \frac{1}{\epsilon_0}Q\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} </math>

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

== See also ==

===Further reading===

http://www.physicsbook.gatech.edu/Point_Charge

http://www.physicsbook.gatech.edu/Electric_Dipole

http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem

==References==

This section contains the the references you used while writing this page

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

https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/

Matter and Interactions, 4th Edition: 1-2

Field of a Charged Ball

2015-12-02T18:43:36Z

Eerwood3:

By Eric Erwood

In this section, the electric field due of sphere charged throughout its volume will be discussed.

==The Main Idea==

In this section, we will focus on a scenario where a sphere has charge distributed throughout the entire object. Calculating the electric field both outside and inside the sphere will be addressed.

===A Mathematical Model===

Step 1: Given a solid charged sphere throughout its volume, the first step is to cut up the the sphere into pieces. As a result, the solid sphere will appear as a series of spherical shells.

Step 2: Relationship between r and R.
Next, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. All the spherical shells appear as point charges at the center of the sphere. As a result, the electric field outside the sphere is a point charge:

<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>
when r>R, and R is the radius of the sphere.

However, when r<R, the observation location is inside some of the shells but outside others.
To find <math>\vec E_{net} </math>, add the contributions to the electric field from the inner shells.
After adding the contributions of each inner shell, you should have an electric field equal to:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2}</math>

We find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

We find that:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

The charge inside the sphere is proportional to r. When r=R,

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} </math>

===A Computational Model===

https://trinket.io/glowscript/48f6efd07a

use this code to see how it looks, change the r value.

==Examples==

===Simple===
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.

Step 1: cut up the sphere into shells

step 2: we know that r<R

Next find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C </math>

===Difficult===
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)
[[File:wikiimage3f.jpg]]

For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud

(s « R in the diagram). (Use the following as necessary: R and ε0.)

Useful equations:

<math> \vec p = \alpha *E</math>

<math> \vec p = Q*s </math>

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find <math> \alpha </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s </math>

<math>\alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 </math>

==Looking Ahead, Gauss's Law==

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

<math>\oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} </math>

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

<math>\vec E * 4\pi*r^2= \frac{1}{\epsilon_0}Q\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} </math>

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

== See also ==

===Further reading===

http://www.physicsbook.gatech.edu/Point_Charge

http://www.physicsbook.gatech.edu/Electric_Dipole

http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem

==References==

This section contains the the references you used while writing this page

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

https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/

Field of a Charged Ball

2015-12-02T18:42:26Z

Eerwood3: /* Difficult */

Claimed by Eric Erwood

In this section, the electric field due of sphere charged throughout its volume will be discussed.

==The Main Idea==

In this section, we will focus on a scenario where a sphere has charge distributed throughout the entire object. Calculating the electric field both outside and inside the sphere will be addressed.

===A Mathematical Model===

Step 1: Given a solid charged sphere throughout its volume, the first step is to cut up the the sphere into pieces. As a result, the solid sphere will appear as a series of spherical shells.

Step 2: Relationship between r and R.
Next, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. All the spherical shells appear as point charges at the center of the sphere. As a result, the electric field outside the sphere is a point charge:

<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>
when r>R, and R is the radius of the sphere.

However, when r<R, the observation location is inside some of the shells but outside others.
To find <math>\vec E_{net} </math>, add the contributions to the electric field from the inner shells.
After adding the contributions of each inner shell, you should have an electric field equal to:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2}</math>

We find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

We find that:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

The charge inside the sphere is proportional to r. When r=R,

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} </math>

===A Computational Model===

https://trinket.io/glowscript/48f6efd07a

use this code to see how it looks, change the r value.

==Examples==

===Simple===
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.

Step 1: cut up the sphere into shells

step 2: we know that r<R

Next find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C </math>

===Difficult===
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)
[[File:wikiimage3f.jpg]]

For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud

(s « R in the diagram). (Use the following as necessary: R and ε0.)

Useful equations:

<math> \vec p = \alpha *E</math>

<math> \vec p = Q*s </math>

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find <math> \alpha </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s </math>

<math>\alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 </math>

==Looking Ahead, Gauss's Law==

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

<math>\oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} </math>

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

<math>\vec E * 4\pi*r^2= \frac{1}{\epsilon_0}Q\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} </math>

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

== See also ==

===Further reading===

http://www.physicsbook.gatech.edu/Point_Charge

http://www.physicsbook.gatech.edu/Electric_Dipole

http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem

==References==

This section contains the the references you used while writing this page

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

https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/

File:Wikiimage3f.jpg

2015-12-02T18:41:41Z

Eerwood3:

File:Wikiimage2.JPG

2015-12-02T18:39:21Z