Chapter 27 Flashcards by austin Ryan

What is the Formula for Gauss law

Electric Flux= ∳EdA=Qin/ε

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at all points inside the volume of a conductor the E field is =

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On a Conductor where does Excess charge go

The Surface

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The E field in a hole of a conductor is=_________ unless a _________ is inside the hole

A) Zero

B) Charge

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The Electric Field at the surface of a charged conductor is _________ to the surface and of magnitude _______ where __________ is the surface charge density at that point.

A) Perpendicular
B) (eta)/(epsilon Naught)
C)(eta)

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What is the Formula for the electric Field at the surface of a charged conductor?

E(surface)=(eta)/(epsilon naught), Perpendicular to the surface

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A Gaussian Cube has 4 Electric Field Vectors going into it and 2 Electric Field vectors going out. All the E Field Vectors have the same magnitude. What is the net charge on this Cube. (+) or (-)

negative

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An electric Field goes through the surface of a plane pointing (theta) degrees above the positive x-axis. What is the Formula you would need to find the Flux in this situation?

Electric Flux=E (perpendicular) A=EAcos(theta)

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What is the formula for the area of a sphere

4(pi)r^2

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What is the Formula for the area of a circle?

(pi)r^2

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A Gaussian surface contains 3 Posotive charges and 2 negative charges and on the outside 6 negative charges. What is the Electric Flux through the surface in terms of q/ε (naught)

+q/ε( naught)

the 6 negative charges on the outside surface have no effect on the flux inside the surface
+3q + (-2q)=+q
see answer above

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Suppose you had a uniformly charged cube. Can you use symmetry alone to deduce the shape of the cube’s electric Field? If so describe the Field Shape. If Not explain why

No, the cube lacks sufficient symmetry to deduce the shape of the Field.

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20N/C of Electric Field goes in through a Triangular surface completely. The Surface is parallel to the electric field. Does the surface enclose a net positive, negative or no charge. Explain.

The Triangular surface doesn’t enclose any charge because 20N/C goes in and 20N/C goes out.

Ein +Eout =(-20N/C)+(20N/C)=0N/C

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A triangle has 30N/C of Electric Field going out of the surface what is the net charge enclosed by the surface.

Ein+Eout=(0N/C)+(30N/C)=30N/C

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If the Electric Field is everywhere tangent to the surface, then Electric Flux is=?

Electric Flux =0

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Which Gaussian surface would allow you to use Gauss’s law to determine the Electric field outside a uniformly charged cube?
A) A sphere whose center Coincides with the center of the charged cube
B) A Cube whose center coincides with the center of the charged cube and that has parallel surfaces
C) A or B
D) neither A or B

D. A cube doesn’t have enough symmetry to use Gauss’s law

The electric field inside a Plastic ball with no charge inside is what? explain your answer.

A Gaussian sphere with a radius smaller than that of the radius of the physical ball is zero inside

The electric field outside of a plastic ball is what? Explain your answer.

∲E⋅dA=EA=Qin/ε the ball has a Constant Area and Electric Field so.
EA=Qin/ε –>E=Qin/Aε —>E=Qin/(4(pi)r^2)ε—> E=(kQin)/r^2

If a closed surface surrounds a dipole, the net flux through the surface is zero. True or False

True

A positive charge Q is located at the center of an imaginary Gaussian cube of sides a. The flux of the electric field through the surface of the cube is Φ. A second, negative charge -Q is placed next to Q inside the cube. What is the net flux through the surface of the cube ?

Outside a spherically symmetric charge distribution of net charge Q, Gauss’s law can be used to show that the electric field at a given distance…

Acts like it originated in a point charge Q at the center of the distribution.

Consider a spherical Gaussian surface of radius R centered at the origin. A charge Q is placed inside the sphere. Where should the charge be located to maximize the magnitude of the flux of the electric field through the Gaussian surface?

The flux does not depend on the position of the charge as long as it is inside the sphere.

Two long straight parallel lines of charge #1 and #2, carry positive charge per unit lengths of λ1 and λ2, respectively. λ1>λ2. The locus (or set) of points where the electric field is zero in this case is…

Along a line between the lines closer to line #2 than line #1

Gaussian surface A and B enclose the same positive charge +Q. The area of Gaussian surface A is three times larger than that of Gaussian surface B. The flux of electric field through Gaussian surface A is…

equal to the flux of electric field through Gaussian surface B

The two spheres in Figure Q27.8 on p 804 surround equal charges. Three students are discussing the situation the situation. Student 1: The fluxes through spheres A and B are equal because they enclose equal charges. Student 2: But the electric field on sphere B is weaker than the electric field on sphere A. The flux depends on the electric field strength, so the flux through A is larger than the flux through B. Student 3: I thought we learned that flux was about surface area. Sphere B is larger than sphere A, so I think the flux through B is larger than the flux through B is larger than the flux through A. Which student is correct?

Student 1

An advantage in evaluating surface integrals related to Gauss's law for symmetric charge distributions is...

The electric field is of constant magnitude on certain surfaces.

If the electric flux through a closed surface is zero, the electric field at points on that surface must be zero. True or false?

False

A conductor is placed in an electric field under electrostatic conditions. Which of the following statements is correct for this situation?

The electric field on the surface of the conductor is perpendicular to the surface; The electric field is zero inside the conductor; A certain fraction of the valence electrons go to the surface of the conductor.

Figure Q27.2 on p 804 shows cross sections of 3-D closed surfaces. They have a flat top and bottom surface above and below the plane of the page. However, the electric is everywhere parallel to the page, so there is no flux through the top or bottom surface. The electric field is uniform over each face of the surface. Which surface encloses negative charge?

Surface C

If the electric flux through a rectangular area is 5.0 N m^2/C, and the electric field is then doubled, what is the resulting flux through the area?

10 Nm^2/C

Stop to think 27.1 on page 783.

A and D. Symmetry requires the electric field to be unchanged if front and back are reversed, if left and right are reversed, or if the field is rotated about the wire's axis. Fields A and D both have proper symmetry. Other factors would now need to be considered to determine the correct field.

Stop to think 27.2 on page 785.

E. A net negative charge.

Stop to think 27.3 on page 791

C. There's no flux through the four sides. The flux is positive 1 Nm^2/C through both the top and bottom because E and A both point outwards.

Stop to think 27.4 on page 794

φb=φe > φa=φc=φd

Stop to think 27.5 on page 799

D. A cube doesn't have enough symmetry to use Gauss's law. The electric field of a charged cube is not constant over the face of a cubic Gaussian surface, so we can't evaluate the surface integral for the flux.