Physics 122B Electricity and Magnetism

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Physics 122B Electricity and Magnetism. Lecture 7 Gauss’s Law and Applications April 09, 2007. Martin Savage. Lecture 7 Announcements. Lecture Homework #2 has been posted on the Tycho system. It is due on Wednesday, April 11 , at 10 PM.

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### Physics 122B Electricity and Magnetism

Lecture 7

Gauss’s Law and Applications

April 09, 2007

Martin Savage

Lecture 7 Announcements
• Lecture Homework #2 has been posted on the Tycho system.It is due on Wednesday, April 11 , at 10 PM.
• On Friday, April 13 we will have Exam 1, covering Knight, Chapters 25-27. You should bring with you to the Exam a good calculator, a Scantron sheet, and one sheet of 8.5”x11” paper on which you have written anything you wish on both sides.

Physics 122B - Lecture 7

Hints on Preparing for Exam 1
• Review Chapters 25, 26 and 27, the lectures on the Web, and the important concepts featured in the Chapter Summaries.
• Fill in identification on Scantron sheet before the Exam.
• Make your 2-sided page of notes and focus on understanding the formulas you write there.
• Make sure you know how to do all of the Tycho homework problems.
• Practice for the exam by working problems at the ends of chapters 25-27.

Physics 122B - Lecture 7

Flux and Charge

Consider the flux through a spherical surface of radius R that surrounds a point charge q located at its center.

This result is independent of how big we choose to make the spherical surface. A very large or a very small surface will have the same flux, which depends only on the size of the charge q at the center of the sphere.

Essentially, q/e0 is the number of field lines attached to charge q, and e0 is the charge producing one unit of flux.

Physics 122B - Lecture 7

Flux is Independent ofSurface Shape and Radius

What about the flux when the surface has an arbitrary non-spherical shape?

We can approximate such a surface by a set of spherical pieces. Summing over these gives Fe=q/e0.

We conclude that the flux through an arbitrary surface always depends only on the charge qenclosed by the surface.

Physics 122B - Lecture 7

Charge Outside the Surface

What about the flux through a surface from any charges not enclosed by the surface?

There is no net flux because entering and exiting fluxes cancel.

Physics 122B - Lecture 7

Multiple Charges

So far we have considered only the flux from a single point charge. What about the case of multiple charges?

Physics 122B - Lecture 7

Gauss’s Law

The previous arguments lead to the conclusion that the correspondence between the charge Qin enclosed by a surface and the net flux Fe through that surface is a general result. It is called Gauss’s Law, and is usually written as:

Johann Carl Friedrich Gauss

(1777 – 1855)

Gauss’s Law is the first of four master equations, collectively called Maxwell’s Equations, that together constitute a “unified field theory” of electromagnetism. In essence, Gauss’s Law says that diverging field lines from a point indicate the presence of an electric charge at that point, and that this charge can be “detected” by surrounding the point with a surface and observing the flux through the surface.

Physics 122B - Lecture 7

This means that the electric field can have only radial components and can depend only on radius. If a concentric spherical Gaussian surface is constructed around the charge distribution, the E-field and dA surface vectors will be parallel, and the magnitude of the E-field will be constant over the surface. Therefore,Fe = ò E·dA = EA, greatly simplifying the flux calculation.Gaussian Spherical Geometry

Spherical distributions of charge have spherical symmetry, so the electric field must also have spherical symmetry.

Physics 122B - Lecture 7

Example:E Inside a Sphere of Charge

What is the electric field Ei at some point x inside a hollow spherical shell of radius R with total charge Q distributed uniformly over the shell?

R

Q

r

x

Put a concentric spherical Gaussian surface through x, so that it has radius r

Then Ei and dA are parallel everywhere on the surface, so Fe=Ei(4pr2)=Qin/e0. But Qin=0 because all the charge lies outside the Gaussian sphere. Therefore, Ei=0.

In other words, the electric field anywhere inside a uniform spherical shell of charge is zero.

Physics 122B - Lecture 7

Example: The E-Field Outside a Charged Spherical Shell

x

What is the electric field Eout at some point x outside a hollow spherical shell of radius R with total charge Qdistributed uniformly over the shell?

r

R

Q

Put a concentric spherical Gaussian surfacethrough x, so that it has radius r>R. From thespherical symmetry of the charge distribution,the electric field Eout on this surface must beradially outward (in spherical coordinates).

Then Eout and dA are parallel everywhere on the surface, so Fe= EoutA = Eout(4pr2)=Qin/e0, with Qin=Q because all of the charge lies inside the Gaussian sphere. Therefore, Eout = (Qin/e0)/(4pr2) = (1/4pe0)Q/r2.

In other words, the electric field anywhere outside a uniform spherical shell of charge Q is the same as the electric field for a point charge Q at the center of the sphere.

Physics 122B - Lecture 7

Summary: The E-Field ofa Spherical Shell of Charge

x

rout

R

Q

x

rin

Physics 122B - Lecture 7

Question 1

(u)

(v)

(w)

(x)

(y)

What is the ranking of the electric fluxes for these systems?

(a) Fx = Fy > Fu = Fv = Fw ;

(b) Fx = Fy = Fu > Fv = Fw ;

(c) Fx = Fy > Fu = Fv > Fw ;

(d) Fv = Fy > Fu = Fw = Fx ;

(e) Fv = Fy > Fu > Fw > Fx .

Physics 122B - Lecture 7

Example: The E-Field Outside a Solid Sphere of Charge

x

What is the electric field Eout at some point x outside a solid sphere of radius R with total charge Qdistributed uniformly in the volume?

r

R

Q

Put a concentric spherical Gaussian surfacethrough x, so that it has radius r>R. From thespherical symmetry of the charge distribution,the electric field Eout on this surface must beradially outward (in spherical coordinates).

Then Eout and dA are parallel everywhere on the surface, so Fe = EoutA = Eout(4pr2)=Qin/e0, with Qin=Q because all of the charge lies inside the Gaussian sphere. Therefore, Eout= (Qin/e0)/(4pr2) = (1/4pe0)Q/r2.

In other words, the electric field anywhere outside a solid uniform sphere of charge Q is the same as the electric field for a point charge Q at the center of the sphere and for a spherical shell of charge Q.

Physics 122B - Lecture 7

Example: The E-Field Inside a Solid Sphere of Charge

What is the electric field Ein at some point x inside a solid sphere of radius R with total charge Qdistributed uniformly in the volume?

R

Q

r

x

Put a concentric spherical Gaussian surfacethrough x, so that it has radius r

Then Ein and dA are parallel everywhere on the surface, so Fe = EinA = Ein(4pr2)=Qin/e0, with Qin=QVG/V0=Q(r/R)3 because only part of the charge lies inside the Gaussian sphere. Therefore, Eout= (Qin/e0)/(4pr2) = (1/4pe0)Q(r/R)3/r2 = (1/4pe0)Qr/R3.

In other words, the electric field anywhere inside a solid uniform sphere of charge Q is proportional to the distance r from the center of the sphere and is equal to zero at the center.

Physics 122B - Lecture 7

Summary: The E-Field ofa Solid Sphere of Charge

x

rout

R

Q

x

rin

Physics 122B - Lecture 7

Example: The E-Field Outside a Long Charged Cylinder

What is the electric field Ecyl at some point x outside a long charged cylinder of radius R and charge per unit length ldistributed uniformly through the cylinder?

x

Put a coaxial cylindrical Gaussian surfaceof length L through x, so that it has radiusr>R. From the cylindrical symmetry of thecharge distribution, the electric field Ecylanywhere on the Gaussian cylinder must be radially outward (in cylindrical coordinates).

Therefore, Ecyl||dA on the curved wall of the Gaussian cylinder and Ecyl^dA on the circular end caps. The enclosed charge is Qin=lL, so Fe = EwallAwall+ EtopAtop+ EbottomAbottom = EcylAwall= Ecyl(2prL)=Qin/e0 Therefore, Ecyl= (Qin/e0)/(2prL) = (1/4pe0)2l/r

Physics 122B - Lecture 7

R

+ + + + + +

x

r

+ + + + + +

Section of long cylinder

Example: The E-Field Insidea Long Charged Cylinder

What is the electric field Ecylat some point xinside a long charged cylinder of radius R and charge per unit length ldistributed uniformly through the cylinder?

Put a coaxial cylindrical Gaussian surfaceof length L through x, so that it has radiusr

Therefore, Ein||dA on the curved wall of the Gaussian cylinder and Ein^dA on the circular end caps. The Gaussian cylinder encloses only part of the charge, so Qin=lLVG/V0=lL(r/R)2, so Fe = EwallAwall+ EtopAtop+ EbottomAbottom = EinAwall= Ein(2prL)=Qin/e0. Therefore, Ein= (Qin/e0)/(2prL) = (1/4pe0)2l(r/R)2/r = (1/4pe0)2lr/R2.

Physics 122B - Lecture 7

Example: The E-Field Outside a Plane of Charge

What is the electric field Eplane at some point xthat is a distance L from an infinite plane with charge per unit area hdistributed uniformly through the plane?

L

x

Put a cylindrical Gaussian surface of length2L and cap area A through x, so that it extendssymmetrically through the plane. From therotational and translational symmetry of the charge distribution, the electric field Eplane anywhere on the Gaussian cylinder must be outward perpendicular to the plane and must not depend on where the axis of the cylinder passes through the plane. Therefore, Eplane^dA on the curved wall of the Gaussian cylinder and Eplane||dA is constant on the circular end caps. The enclosed charge is Qin=hA, so Fe = EtopA + EbottomA = A(Etop+ Ebottom)= 2EplaneA=Qin/e0 = hA/e0. Therefore, Eplane = h/2e0.

Physics 122B - Lecture 7

Conductors and Gauss’s Law

The electric field must be zero inside any conductor in electrostatic equilibrium because charges are free to move in a conductor, and any internal field would cause the mobile charges to rearrange themselves until the field disappeared.

Now suppose that we place a Gaussian surface just inside the “skin” of a conducting object. The flux through this Gaussian surface is Fe= 0 because E= 0. But Fe= Qin/e0, so Qin= 0 also. Therefore, all of the charge present in a charged conductor resides on the outside surface of the conductor.

Physics 122B - Lecture 7

The Surface Field of a Conductor

Suppose that the electric field at the surface of a conductor had a component tangent to the surface. Then mobile electric charges at the surface would feel a force and would move in response to the field, until equilibrium was established and the tangential field vanished. Therefore, the field at the surface of a conductor must be perpendicular to the surface, with no tangential components.

Suppose that a charged conductor has local surface charge density h. Put a perpendicular cylinder of end-cap area A through this region of the surface. The field on the inner end cap is zero, and the field is perpendicular to the area vectors on the curved wall of the cylinder, so Fe =EsurfA= Qin/e0 = hA/e0. Therefore, Esurf = h/e0. (Note that this matches the surface field of a sphere.)

Physics 122B - Lecture 7

Charges and FieldsInside a Conductor

Suppose there is a cavity in the interior of a conductor. Can any of the net charge of the conductor reside on the surface of the cavity?

Place a Gaussian surface in the conductor just outside the cavity surface. There E = 0 so Fe=0, so Qin= 0. Therefore, there is no charge or field within or at the surface of the cavity.

Now suppose we force the issue by placing a point charge q in the interior cavity of a neutral conductor. Using the same Gaussian surface, Qin= 0 as before. This means that a compensating charge -q must be present on the interior cavity surface, to cancel the field of the interior point charge, and to compensate for that, a charge qmust be present on the outer surface of the conductor.

Physics 122B - Lecture 7

Screening E-Fields

Suppose that we wish to create a region where the electric field is zero within a region containing a strong electric field.

All we have to do is surround the region with a neutral conductor, and the conducting properties will ”screen” the region, insuring an electric field of zero. The charges on the outside of the inserted conductor will distribute themselves so as to “stop” the incoming field lines on one side of the conductor and “restart” them on the other side.

The “price” paid for such screening is that the exterior electric field may be very distorted by the presence of the inserted conductor.

Physics 122B - Lecture 7

End of Lecture 7
• Lecture Homework #2 has been posted on the Tycho system.It is due on Wednesday, April 11 , at 10 PM.
• On Friday, April 13 we will have Exam 1, covering Knight, Chapters 25-27. You should bring with you to the Exam a good calculator, a Scantron sheet, and one sheet of 8.5”x11” paper on which you have written anything you wish on both sides.

Physics 122B - Lecture 7