Download presentation

Presentation is loading. Please wait.

Published byThomas Ortega Modified over 3 years ago

1
A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is -3 Q, and it is insulated from its surroundings. Derive expressions for the electric field magnitude E in terms of the distance from the center r for the regions r b. a b -3Q +Q Bold type denotes vector quantities

2
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9

3
A : IA B : II & IVB C : IIIC D : III & IVD 1.Which of the following physics principles should one use to solve this problem? I.Amperes Law II.Faradays Law of Induction III.Gausss Law IV.Superposition of Electric Fields

4
This law deals with magnetic fields produced by electric current. Choice: A Incorrect

5
Faradays law deals with the time rate of change of magnetic flux, so this is not applicable to our situation. On the other hand, the principle of superposition of electric fields is very helpful here. Considering the contributions to the electric field from each charge will make this task easier to evaluate. Choice: B Incorrect

6
We can use this law to solve for E as we exploit symmetry, but another physics principle will also be helpful. Choice: C Incorrect

7
Choice: D Correct Gausss law and the principle of superposition of electric fields are very helpful here. Considering the contributions to the electric field from the charge in each region will make this task easier to evaluate.

8
2. Which statement correctly describes Gausss Law? A : The total electric field through a closed surface is equal to the net charge inside the surface.A B : The total electric flux through a closed surface is equal to the total charge inside the surface divided by the area.B C : The total electric flux through a closed surface is equal to the total charge inside the surface divided by o (permittivity of free space).C

9
Gausss Law relates the electric flux through a surface to the total charge enclosed by the surface. Choice: A Incorrect

10
Choice: B Incorrect According to Gausss Law, the total electric flux through a closed surface is equal to the total charge inside the surface divided by o, not by the area.

11
Choice: C Correct In mathematical form, Gausss Law is expressed as:

12
3. What is a convenient Gaussian surface for this system? A: circleA B: cubeB C: sphereC

13
Choice: A Incorrect The system is three dimensional.

14
The magnitude of the electric field is not the same at all points on a cubical surface. Choice: B Incorrect

15
Choice: C Correct This is a good choice, because the system shows spherical symmetry.

16
Please get out a piece of scratch paper and sketch the charge configuration of the situation described in the problem statement. For now, only draw the arrangement due to the charge +Q inside the cavity. Draw your Gaussian surface with a radius r

17
Example: Your sketch should look something like this: a r+Q Q Q 1. A negative charge of magnitude -Q is induced uniformly around the inner surface of the cavity. 2. The negative charge is drawn to the inner surface and a positive charge remains on the outer surface of the conducting shell. Gaussian surface

19
This is a correct expression, but it is not the only one. Choice: A Incorrect

20
This is a correct expression, but it is not the only one. Choice: B Incorrect

21
Check the units. This is an expression for electric field. Choice: C Incorrect

22
Both these expressions are correct. Choice: D Correct Since the electric field is perpendicular to the infinitesimal area dA, the magnitude of the electric field E can be pulled out of the integral. Mathematical expression of Gausss Law: The total charge enclosed in our Gaussian surface is only +Q, therefore:

23
One of these expressions is not correct. Try Again. Choice: E Incorrect

24
5. Which one of the following expressions describes the electric field E in the region r

25
Choice: A Incorrect The total charge enclosed in the Gaussian sphere is +Q.

26
Choice: B Correct From the previous question we have: Simple algebra shows that:

27
Choice: C Incorrect Since our Gaussian surface has a radius r less than a, the only charge that is enclosed is the point charge +Q.

28
6. For electrostatic equilibrium, the electric field inside the conductor (metal) is which of the following? A: uniform but non-zeroA B: zeroB C: non-uniformC

29
Choice: A Incorrect This is in violation of the equilibrium condition.

30
Choice: B Correct This implies equilibrium.

31
Choice: C Incorrect The charges are uniformly distributed.

32
Please make another sketch of the charge configuration of the spherical shell, this time only consider the -3Q charge that is placed on the outer surface of the conductor.

33
Example Q The charge -3Q distributes uniformly on the surface of the conductor, making the field everywhere inside of the conductor equal to zero. Notice that there is no charge enclosed if the gaussian surface is placed anywhere inside of the surface of the conductor.

34
Considering a superposition of the electric fields produced by the point charge +Q and the surface charge -3Q will help us find expressions for the electric field in the regions a b. Sketch the charge configuration of the spherical shell considering all sources of electric field (combine your two previous drawings).

35
Superposition of electric fields and charge summation += Q + -Q +Q -Q -3Q +Q The red arrows depict electric field lines. Notice that there are more field lines in the second drawing pointing inwards than there are in the first pointing outwards.

36
7. Add a Gaussian surface to your drawing with a radius a

37
Choice: A Correct The electric field inside of a conductor is always equal to zero. r a b

38
Choice: B Incorrect Remember that the electric field inside of a conductor is always zero. Your Gaussian surface should enclose more than just the point charge +Q.

39
Remember that the electric field inside of a conductor is always zero. Your Gaussian surface should enclose more than just the induced charge -Q on the inside of the shell. Choice: C Correct

40
8. What is the net charge enclosed by a Gaussian sphere in the r>b region? A: +QA B: -3QB C: -QC D: -2QD

41
Choice: A Incorrect b r Gaussian surface The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

42
Choice: B Incorrect b r Gaussian surface The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

43
Choice: C Incorrect b r Gaussian surface The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

44
Choice: D Correct b r Gaussian surface The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

45
9. Which one of the following is the expression for the electric field E in the region r>b? AA: (radially outward) CC: (radially inward) BB: (radially outward) DD: (radially inward)

46
The total enclosed charge is not +Q. Also, field lines from a negative source charge are directed inward. Please check your sketch and try again. Choice: A Incorrect

47
Choice: B Incorrect Field lines from a negative source charge are directed inward.

48
Choice: C Correct Reasoning:

49
Choice: D Incorrect This is the total flux through the Gaussian sphere.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google