1 AP Physics C E & M Gauss's Law 20160109 www.njctl.org 2 - - PowerPoint PPT Presentation

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AP Physics C ­ E & M

Gauss's Law

2016­01­09 www.njctl.org

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Gauss's Law

Click on the topic to go to that section

• Electric Flux
• Gauss's Law
• Sphere
• Infinite Rod of Charge
• Infinite Plane of Charge
• Electrostatic Equilibrium of Conductors

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Electric Flux

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Electric Flux

An Electric Field, , represented by the Electric Field lines below, passes through a rectangular area of space, with cross sectional area, A. The field is perpendicular to the area.

Define Electric Flux as the strength of the Electric Field times the area through which it passes: Visually, we use the number

• f field lines to represent the

Electric Field strength ­ so, the more lines passing through an area of space, the greater Electric Flux.

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Electric Flux

But what if the Electric Field is not perpendicular to the area of space that we're interested in finding out the Electric Flux? Find the Electric Flux through the slanted area of the below shape.

The Electric Field makes an angle θ with the vector dA which is normal (perpendicular) to the light blue surface area. A is the area of the slanted side and A1 is the area of the vertical side.

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Electric Flux

The number of field lines through surfaces A1 and A is the same. The widths of each surface are the same, and their lengths are two sides of a triangle where A has the length of the hypotenuse. Since the flux through both surfaces is the same: The flux is dependent on the number of field lines, and proportional to the angle that the normal to the surface makes with the field.

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Electric Flux

Not all surfaces are ramps. Let's see how calculus will help in calculating the flux through an arbitrary shape. The flux through any area element ΔAi is: To find the total flux through this surface, shrink ΔAi to almost zero and sum all the ΔΦEi's over the entire surface: Note the different symbol for the integral ­ that circle means you are integrating over an entire surface ­ it is called a surface integral.

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1 A rectangular loop is held perpendicular to a uniform Electric Field. At what angle to its normal does the loop have to be turned through so the electric flux is one half

• f its original value?

A

15

B

30

C

45

D

60

E

90

• Answer

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2 A rectangular loop is held parallel to a uniform Electric Field and is turned through an angle of 350 clockwise. What is the Electric Flux through the loop after it is turned? A 1.1 x 104 Nm2/C B 7.8 x 103 Nm2/C C 1.1 x 108 Nm2/C D 7.8 x 107 Nm2/C E Zero

Answer

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Electric Flux through a Closed Surface

A closed surface is a shape that has an inside and an outside. And you can't get from the inside to the outside without passing through the surface. Not the most profound of physics definitions, but it leads to a very powerful method for finding electric fields ­ Gauss's Law. For Electric Field lines passing through a closed surface, we define a sign convention:

• Field lines going from the outside to the inside of a closed

surface are given a negative sign.

• Field lines going from inside to the outside of a closed surface

are given a positive sign. The net Electric Flux through a closed surface is then equal to the number of field lines leaving minus the number of lines entering.

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Charge and Electric Flux

Here are three identical boxes with different combinations of charges within their closed surfaces. Each charge has the same magnitude and generates an electric field represented by the red field lines. What is the Electric Flux through each box?

+

Box A

­

Box B

­

+

Box C

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Charge and Electric Flux

Box A: The electric field lines point out of the box, therefore it has an outward electric flux (positive). Box B: The electric field lines point into the surface of the box, therefore it has an inward electric flux (negative). Box C: Has a net charge of zero within the box, and since the flow is the same but in opposite directions for each charge, the electric flux is zero.

+

Box A

­

Box B

­

+

Box C

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An empty box is placed in an external Electric field. The amount of charge enclosed by the box is zero, and the net electric flux is zero because the number of field lines entering is equal to the number of field lines leaving.

Charge and Electric Flux

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+q

+2q

If the electric field doubles, then the number of field lines representing the field through the surface will also double. From this we can conclude that the net electric flux is directly proportional to the enclosed charge.

Charge and Electric Flux

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Charge and Electric Flux

A positive charge is surrounded by two concentric spheres ­ what is the relationship of the electric flux through each sphere?

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Charge and Electric Flux

The Electric Flux through each sphere is the same. As the distance from the charge increases, the Electric Field decreases as 1/r2, while the surface area that encloses the field lines increases as a factor of r2. Since Electric Flux is a product of the field strength and the surface area, these factors cancel out. The Electric Flux is independent of the size of the enclosing geometry.

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3 A positive charge is enclosed by a spherical surface. What direction is the electric flux relative to the surface? A Tangent. B Always perpendicular. C From the outside to the inside of the surface. D From the inside to the outside of the surface. E The lines do not pass through the surface.

Answer

D

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4 A charged particle is enclosed within a rectangular box and produces an Electric Flux of Φ through the box. What is the Electric Flux if the dimensions of the box are tripled? A 1/9 Φ B 1/3 Φ C Φ D 3 Φ E 9 Φ

Answer

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Gauss's Law

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Gauss's Law

Electric Flux will now be used to derive Gauss's Law. Start with a positive charge, q, and surround it with a hypothetical spherical surface ­ called a "Gaussian surface." This surface has no physical meaning ­ it's being used for its symmetry properties, and is typically shown as a dashed line or a shaded surface. Gaussian surface

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Gauss's Law

For any segment on the surface of the sphere, ΔAi, its normal vector, dA

i and the Electric Field,

E are parallel. Furthermore, E is constant everywhere on the surface. The flux through this segment is: The total flux through the Gaussian surface is then:

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Gauss's Law

The magnitude of E at a distance r from a point charge is: Substitute this value in the flux equation found on the previous slide: This is one of the reasons that ε0 is used instead of κ ­ we get rid of 4π in Gauss's Law.

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Gauss's Law

One more step ­ to show that works for any closed surface surrounding a charge. All three surfaces have the same Electric Flux through them ­ visually, the same amount of Electric Field lines go through each surface. Since they all enclose the same charge, the above equation works, regardless of the shape of the enclosing surface.

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Gauss's Law

If there is more than one charge enclosed by the surface, you just need to add the charges algebraically and Gauss's Law applies. This is a powerful equation, but several assumptions are required in order to use it to solve problems using simple

• integrals. Here's the list:
• The Electric field is constant (which could also equal zero) at

certain segments of the chosen Gaussian surface.

• The Electric field is perpendicular to the surface so that
• The Electric field is parallel to the surface so that .

. or,

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Charge Density

When finding the Electric field due to a charge distribution, the following definitions (from last chapter) will be useful. Linear charge density: Surface charge density: Volume charge density:

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5 A charge of magnitude q, is at the center of a sphere of radius r. What is the electric flux at the surface of the sphere? A 5.6x105 Nm2/C B 6.5x105 Nm2/C C 7.3x105 Nm2/C D 1.1x106 Nm2/C E 1.3x106 Nm2/C

Answer

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6 You are trying to calculate the Electric Field outside a positively charged metal sphere of radius R. The Gaussian surface should be what kind of geometrical

• bject?

A Circle B Cylinder C Cube D Sphere E Rectangle

+ + + + + + + + + + + R

Answer

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7 You are trying to calculate the Electric Field outside a positive line of charge. The Gaussian surface should be what kind of geometrical object? A Circle B Cylinder C Cube D Sphere E Rectangle

Answer

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Sphere

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Symmetric Charge Distributions

There are three common symmetric charge distributions that are used to illustrate Gauss's Law. They are:

• Sphere (both conducting and insulated)
• Line of charge (conducting and insulated)
• Infinite plane of charge

We'll start with the Sphere.

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Insulated Sphere

An insulated sphere of radius, R, is uniformly positively charged with a volumetric charge density, ρ and total charge Q. Find the Electric Field inside and outside the sphere.

R Q r

Start with the field outside the

• sphere. Construct a spherical

Gaussian surface of radius, r. The Electric Field is perpendicular and constant everywhere on the surface, so Gauss's Law is appropriate.

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Insulated Sphere

R Q r

The surface integral of the sphere is merely the equation for the surface area of a sphere!

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Insulated Sphere

R Q r

Calculate the Electric Field within the

• sphere. Construct a Gaussian surface

within the sphere. Only the enclosed charge contributes to the field, so Q can't be used. We have to find the part

• f Q that exists within the Gaussian
• surface. Assume a uniformly

distributed charge and take the ratio of the two spheres times the total charge:

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Insulated Sphere

R Q r

We have a two part solution for the Insulated Sphere's Electric Field: for r > R for r < R

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Insulated Sphere

R Q r

For r = R, both expressions give the same value for the Electric

• Field. That's good.

For r > R, the Electric Field due to an insulated sphere is the same as a point charge located at the center of the sphere. A sphere acts as a point charge with the field measured from the center of the sphere. for r > R for r < R

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Conducting Sphere or Thin Shell of Charge

R Q r R Q r

For both of these cases, the Electric Field outside is the same as for the Insulated sphere, as the enclosed charge is Q.

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Conducting Sphere or Thin Shell of Charge

R Q r R Q r

However, since the charge for a conducting sphere resides on the surface, when we draw a Gaussian surface right inside the sphere, we get an enclosed charge of zero ­ just like a surface within the thin shell, hence, zero Electric Field for both cases.

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Conducting Sphere or Thin Shell of Charge

R Q r R Q r

Again, there is a two part solution for this case. for r > R for r < R

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8 A metal sphere of radius R is given a positive charge, q. What is the electric field at a distance r, where r > R? A B C D E

Answer

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9 A metal sphere of radius R is given a positive charge, q. What is the electric field at a distance r, where r < R? A B C D E

Answer

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10 A metal sphere of radius R is given a positive charge, q. What is the electric field at a distance r, where r = R? A B C D E

Answer

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11 Which is the graph of the Electric Field for a charged conducting sphere of radius R?

A B C D

R

E Emax

E

R

E Emax

R

E Emax

R

E Emax

R

E Emax

Answer

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12 An insulated sphere of radius R has a uniform charge distribution of magnitude Q. Which is the graph of its Electric Field?

A B C D

R

E Emax

E

R

E Emax

R

E Emax

R

E Emax

R

E Emax

Answer

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13 A nonconducting sphere of radius R has a uniform charge distribution of magnitude Q. What is the electric field at a distance less then R?

A B C D E R

Answer

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Infinite Rod of Charge

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Infinite Rod of Charge

Consider an infinite rod of charge of linear density λ and radius R. Find the Electric field inside and outside the rod. Construct a Gaussian surface taking advantage of the cylindrical symmetry of the wire ­ at a distance r from the wire, the Electric Field is constant and perpendicular to the surface of the cylinder. Since the wire is infinite, we don't have to worry about the ends of the wire ­ where there would be a parallel component of the Electric Field and Gauss's Law can't be applied.

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Infinite Rod of Charge

Surface area of the curved part of the cylinder This works whether the rod is insulated or a conductor. Next, find the Electric Field within the rod.

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Infinite Rod of Charge

If the rod is a conductor, then the Electric Field within the rod is zero ­ there is no enclosed charge. For an insulator of volume charge density ρ, we'll construct a Gaussian surface of length l and radius r within the rod. Since the rod is of infinite length, there is no Electric Field through the end caps of the Gaussian cylinder. The Electric Field is uniform and perpendicular to the curved surface.

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Infinite Rod of Charge

Enclosed charge is the volumetric charge density times the volume of the cylinder. Surface area of the cylinder.

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Infinite Rod of Charge

Enclosed charge is the volumetric charge density times the volume of the cylinder. Surface area of the cylinder.

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Gaussian Surface

14 There is an electric charge distributed uniformly over an infinitely long metal wire of radius R. What is the electric field at a distance r (r > R) from the center of the wire?(The charge per unit of length is .)

A B C D E

Answer

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Infinite Plane of Charge

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Infinite Plane of Charge

Here is an infinite plane (pretend the plane extends infinitely in the x and y direction)

• f positive charge with an

Electric Field pointing out of the

• page. Find the magnitude of

the field. The charge density on the plane is σ. Rotate the plane sideways so it looks like an infinite line.

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Infinite Plane of Charge

Draw a Gaussian surface ­ a cylinder ­ which is composed of two end caps, each with a surface area of A, and a curved surface (open tube). End Cap Curved surface End Cap

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Infinite Plane of Charge

Every point on the curved surface is perpendicular to the Electric field generated by the positively charged plane of charge. The curved surface does not contribute anything to the net Electric Field. End Cap Curved surface End Cap

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End Cap Curved surface End Cap

Infinite Plane of Charge

There are two end caps, each with the Electric Field perpendicular to the surface area of the cap: The flux through one cap is . Since there are two caps, the total flux through the Gaussian surface of the two caps and the curved surface is .

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Infinite Plane of Charge

The charge enclosed by the Gaussian surface is equal to the area of the infinite plane that is subtended by the end caps, A, times the surface charge density, σ. Putting this together with the flux determined on the previous page: The Electric Field is constant near the infinite plane. End Cap Curved surface End Cap

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Symmetry Summary

for r > R for r < R Gauss's Law was used to solve for three symmetries, where R is the radius of the sphere or rod.

• Spherical
• Rod
• Infinite plane

for r > R for r < R These solutions are for uniform distributions of charge (insulators). If the sphere or line of charge is a conductor than E = 0 for r < R.

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Answer

15 Describe the Electric Field dependence on r as Gauss's Law is used to solve for a sphere, a rod and an infinite plane of charge.

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Electrostatic Equilibrium of Conductors

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Electrostatic Equilibrium of Conductors

Charges are free to move within the body of a conductor ­ and there are a great number of these charges. Electrostatic Equilibrium occurs when there is no net movement of these free charges. A conductor in electrostatic equilibrium has the following properties:

• The Electric Field within the conductor is equal to zero.
• Any excess charge on an isolated conductor (non grounded)

resides on its surface.

• The Electric Field outside the conductor is perpendicular to the

surface and equal to the charge density divided by the electrical permittivity.

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Zero Electric Field

The Electric Field within the conductor is equal to zero. If there were an Electric Field within the conductor, then there would be a force on the charges, which would result in a net movement of the charges ­ hence, no equilibrium. But what if an external Electric Field is applied to the conductor? ?

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Zero Electric Field

The charges will rearrange themselves ­ the negative charges will be attracted to the left side of the conductor, leaving an equal positive charge on the right side. This creates an opposing field to the applied field. The charges will move until this field is equal and

• pposite ­ then the charges will once again be in electrostatic
• equilibrium. This process takes nanoseconds.

+ + + + + + + ­ ­ ­ ­ ­ ­ ­

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Excess Charge on Surface

Gauss's Law is used to show that any excess charge on a conductor must reside on its surface. Create a Gaussian surface just inside any arbitrarily shaped conductor. The Electric Field inside a conductor in electrostatic equilibrium is zero. By Gauss's Law, if the Electric field is zero, then the net enclosed charge is zero. The Gaussian surface can be drawn infinitely close to the surface of the conductor ­ thus any free charge must reside on the surface of the conductor.

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End Cap Curved surface End Cap

External Field perpendicular

The external Electric Field is perpendicular to the surface. Construct a Gaussian cylinder that pokes through the surface. We'll reuse the cylinder that was used for an infinite plane of

• charge. If the cylinder is small enough, the surface looks flat.

Unlike the infinite plane, there is no Electric Field pointing to the left of the Gaussian surface because the Electric Field equals zero within a conductor.

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End Cap Curved surface End Cap

External Field perpendicular

There is no Electric Field parallel to the surface ­ if there was, a force would be exerted on the charges and they would move ­ they would not be in electrostatic equilibrium. Thus, the Electric Field is perpendicular and pointing outside the surface at every point.

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+ + + + + + + + +

+Q

+

Charge Distribution within a Conductor

Take a conductor with charge +Q. The entire charge resides on the surface as shown below.

+ + + + + + + + + +

+Q Now punch a hole through the conductor, creating an open

• cavity. The charge still is on

the outside surface as shown by the blue Gaussian surface drawn within the conductor ­ there is no Electric Field within the conductor, hence no enclosed charge.

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+ + + + ++ +++ ++ ++ + + + + + + +

+Q

­ ­ ­ ­ ­

­­ ­ ­

­Q

Charge Distribution within a Conductor

Place a charge, +Q within the cavity. What happens to the charge distribution? This charge will generate an electric

• field. But, an electric field cannot

exist within a conductor at electrostatic equilibrium. The charges within the conductor will redistribute themselves such that there will be an enclosed charge of zero within the Gaussian surface. This results in an induced charge of ­Q on the inner surface of the

• conductor. But where did this come

from?

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+ + + + ++ +++ ++ ++ + + + + + + +

+Q

­ ­ ­ ­ ­

­­ ­ ­

­Q +2Q

Charge Distribution within a Conductor

Conservation of Charge requires that the conductor maintains its total

• riginal charge of +Q. Since ­Q is on

the inside, +2Q will be distributed on the external surface ­ summing to +Q.

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16 Which of these are properties of a conductor in electrostatic equilibrium? A I B II C III D I and II E II and III

I ­ The Electric Field is a constant, non zero value within the conductor. II ­ Excess charge resides on the surface of the conductor. III ­ The Electric field just outside the conductor is perpendicular to the surface. Answer