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Gauss's Law©2011 by Bryan Pflueger

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Charge and Electric FluxIf an unknown charge is placed within a box how can you determine the magnitude and sign of that charge?

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You have learned before that every charge produces an electric field. Now if we take a test charge of known magnitude and sign and move it around the box we can map the electric field of the charge on the inside. By knowing the force exerted on the test charge at each point we can determine both the unknown charge's magnitude and sign.

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Charge and Electric FluxA variety of different cases can be shown for charges within a box. The magnitude of each charge in the examples is Q.

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Box A

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Box B

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Box CVelocity is a vector and deals with the motion of objects. Electric Field is a vector as well, but despite the fact that nothing is actually moving in the Electric Field we consider it to "flow" and this is called the 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.

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Charge and Electric Flux

An empty box is held a distance away from a positively charged plate. The amount of charge enclosed by the box is zero, and the net electric flux is zero because the rate at which it is entering on one side is the same as the rate at which it is leaving.

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Charge and Electric Flux

+q +2q

If the electric field doubles then the "flow" of it through the surface will also double. From this we can conclude that the net electric flux is directly proportional to the enclosed charge.

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+q

Charge and Electric Flux

+q

If the same charge were enclosed in another box with twice the dimensions how would this effect the electric flux? Well if the distances were to double the net electric filed lines at this distance would be E/4 and since the electric flux is dependent on rate at which it flows out of the surface. We find that the area is now 4 times greater. This shows that the electric flux due to a charge enclosed within any shape will be independent of its size.

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1 In what direction does a positive charge's electric field point?

A Outward

B Inward

C Tangent to the surface of the charge

D A positive charge does not generate an electric field

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2 A charged paticle is enclosed within a rectangular box and has a electric flux of , what would be the electric flux if the dimensions of the box were tripled?

A 3 times greater

B 1/3 of its orignal value

C 1/9 of its original value

D Remains the same

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Charge and Electric FluxTo calculate the electric flux lets first return to the idea that the electric field is flowing like water. If the water flows at a constant rate of v through an area of A, what is the volume flow rate?

If the area through which the water flows through is tilted at an angle # , the above equation can be rewritten as:

or as (This is the dot product)

In this Equation we consider the velocity to be the Electric Field and the volume flow rate to be the electric flux. Therefore the equation can be written as:

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Charge and Electric Flux

This equation can be used to solve for the electric flux through any flat surface with a constant Electric Field. But what if the Electric Field varied or if the shape had a curved surface?

To account for this we would have to slice the surface area into little pieces and use the electric field at each point. This can be represented as:

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3 A rectangular loop is held perpendicualr to a uniform Electric Field. At what angel does the loop have to be turned through so the electric flux is one half of its original value?

A 15

B 30

C 45

D 60

E 90

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o

o

o

o

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4 If a rectangular loop is held parallel to a uniform electric field and turned through an angel of 35 degrees clockwise, what is the electric flux through the loop?

A

B

C

D

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Gauss's LawGauss's law states "the total electric flux through any closed surface is proportional to the net electric charge inside the surface."Before we stated that we would use the alternative representation for the electric field equation and in Gauss's Law you will see why this is useful.

In previous slides we proved that the electric flux is equal to the Electric Field times the area through which it "flows".

If we solve the electric flux for a sphere of radius R we can derive Gauss's Law. A is the surface area of a sphere which is 4#R2.

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Gauss's Law

Guass's Law works for any imaginary closed surface. These surfaces will be refered to as Gaussian surfaces. Now what if more then one charge is enclosed within the Gaussian Surface? Each charge will generate its own electric field so we use E as the net electric field and q is rewritten as Qenclosed because we have to take into account each charge enclosed within the surface. We now rewrite Gauss's Law as:

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5 If a charge of magnitude q is at center of a sphere of radius r what is the electric flux through its surface?

A

B

C

D

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6 A metal sphere of radius R is given a positive charge q and we are trying to find the electric field at a distance greater then R. What is the Gaussian surface that we have to use?

A Circle

B Sphere

C Cylinder

D Box

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+R

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7 A metal sphere of radius R is given a positive charge q. What is the electric field at a distance greater then R?

A

B

C

D

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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 less then R.

A 0

B

C

D approaching #

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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 equal to R.

A

B

C

D

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10 What would be the graph of the electric field for a charged metal sphere?

A B

R

E

Emax

C D

R

E

Emax

R

E

Emax

R

E

Emax

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11 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

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12 A nonconducting sphere of radius R has a uniform charge distribution of magnitude Q. What 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

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13 There is an electric charge distributed uniformly over an infinitely long metal wire of radius r. What is the electric field produced by this charge distribution? (The charge per unit of length is )

A

B

C

D

E

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Charges on Conductors

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(a) (b) (c)

(a) If a conductor has a charge of q, then the charge on the surface of that conductor will also be q, because in a conductor all of the charge moves out towards the object's surface.

(b) If no charge resides on the cavity's wall the net charge is zero.

(c) If a charge with magnitude q is placed within the cavity, the net charge on the surface of the cavity is -q and the overall charge on the surface of the conductor is 2q.

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