General Physics – PH 213
General Physics – PH 213 Name:____________
Midterm I (Ch 25 - 27)
October 24, 2008
• Exam is closed book, closed notes. Use only your note card.
• Write all work and answers in the color papers provided.
• Show all your work and explain your reasoning (except for true/false or multiple choices ones)
• Partial credit will be given. No credit will be given if no work is shown (except for true/false or multiple choices ones)
Part I – True or False: For questions 1 – 8, state whether each statement is true or false.
1. (True) The electric field of an infinite sheet of charge is the same everywhere above the sheet.
2. (True) Any charge q can be expressed as [pic] where n is an integer and e is the charge of an electron.
3. (True) The An electron moving in +x direction is slowed down when it enters the field [pic].
4. (False) If the net charge on a conductor is zero, its surface charge density must be uniform.
5. (False) If the net flux through a closed surface is zero, then there can be no charge within that surface.
6. (True) Units of electric fields are newtons per coulomb, N/C.
7. (False) Inside a conductor the electric field is always zero.
8. (True) A charged insulator and uncharged metal always attract each other electrostatically when brought near each other.
Part II – Multiple choice questions: For questions 9 – 16 select the one alternative that best completes the statement or answers the question.
9. The units of electric dipole moments are
A) N/m B) C2(m C) N/C D) C2/m E) m(C
10. A positive point charge is placed at the center of a spherical conducting shell with radii Ri and Ro as shown. Which of the following statements is true for the electric field vector?
A) E[pic] for r < Ri and E = 0 for r > Ro .
B) E[pic] for r > Ro and E = 0 for r < Ri .
C) E[pic] for r < Ri and for r > Ro .
D) E[pic] for Ri < r < Ro .
E) The E is zero everywhere.
11. When a negatively charged conductor touches a neutral conductor, the neutral conductor will
A) gain protons. D) lose electrons.
B) lose protons. E) gain electrons.
C) stay neutral
12. Particle A of charge +2q is located at the origin, and particle B of charge +q is located on the positive x-axis. The electric force on the particle B due to particle A is represented by vector F. Which vector represents the force on particle A due to particle B? +2q +q F
A B
A) F E) 1/2 F
B) -F F) -1/2 F
C) 2F G) The force is zero.
D) -2F
13. A charged particle of mass m is suspended between the horizontal plates of a charged capacitor. Which of the following statements is not true?
A) The magnitude of the electric force on the
particle is equal to its weight.
B) The electric field between the plates
point downward.
C) The plates are at different potentials.
D) The particle is charged positively.
14. A uniform spherical shell of charge of radius R surrounds a point charge at its center. The point charge has value Q and the shell has total charge -Q. The electric field at a distance R/2 from the center
A) is zero
B) does not depend on the charge of the spherical shell
C) is half of what it would be if only the point charge were present
D) is directed inward if Q>0.
15. A conducting rectangular plate is connected to ground through a switch. The switch is initially closed. A negative charge Q is brought near the plate as shown, but the charge is not brought in contact with the plate. While the charge is near the plate, the switch is opened and then the charge is removed. What is now the charge state of the plate?
A) It is negatively charged
B) It is positively charged.
C) It is uncharged.
D) It can be any of the above depending –Q
on the initial charge on the plate before
the charge –Q was brought nearby.
16. A charge Q is uniformly distributed throughout a nonconducting sphere of radius R. The charge density in the sphere is
A) [pic] C) [pic]
B) [pic] D) [pic] E) Not given
Part III – Questions and Problems
Please explain or show your work clearly for each of the following questions or problem.
17. The figure shows four spheres, each with charge Q uniformly distributed through its volume.
A) Rank the spheres according to their volume charge density, greatest first.
Uniform volume charge density is given by: [pic], and since charge is the same for all spheres, the larger the volume, the smaller the charge density ρ. Therefore:
ρa > ρb > ρc > ρd
B) The figure also shows a point P for each sphere, all at the same distance from the center of the sphere. Rank the spheres according to the magnitude of the electric field they produce at point P, greatest first
Take a spherical Gaussian surface passing over point P for each sphere (r = OP), which will be the same size for all spheres. The electric field value is related to the enclosed charge inside the Gaussian surface. For a and b, the enclosed charge is the same (all of Q), and c contains more charge than d (higher charge density) but less than b or a. So: Ea = Eb > Ec > Ea
18. A positively charged rod is held near, but not touching a neutral metal sphere.
A) Add plusses and minuses to the figure to show the charge distribution on the sphere.
The positive charge will induce charges on the sphere as shown.
B) Does the sphere experience a net force? If so, in which direction. Explain
Yes it does. There is an attractive force between the rod and front side of the sphere, and repulsion with the far end. Since the attractive force is larger (due to shorter distance), there is a net charge on sphere to the right.
19. A point charge +Q is placed at the center of the conductors. Find the induced charges for all surfaces (i.e inside1, I1, outside1, O1, and inside2, I2 and outside2, O2).
Since inside the conductor, E= 0, the electric flux through a Gaussian surface taken inside the first conductor is zero. This in turn means the enclosed charge by this surface must add up to zero. Therefore: qenc = +Q + QI1 = 0 ( QI1 = –Q. Now since inside conductor is neutral, total charge on its inside and outside surfaces must add up to neutral.
Hence: QI1 + QO1 = 0 ( QO1 + (– Q) = 0, ( QO1 = +Q
Using another Gaussian surface inside the 2nd conductor, and using same arguments as above, we conclude: QI2 = – Q, QO2 = +Q
20. An 8.85 ( 10-12 C point charge sits at the origin of a coordinate system. What is the flux of electric field through a 10 cm tall right circular cylinder of diameter 6.0 cm centered on the origin?
[pic]
We do not need to calculate the integral for the flux, since it can be calculated from the right side as well:
[pic] so [pic]
21. Five +1.0 nC point charges are equally spaced around a semicircular arc of 10 cm radius as shown. An electron sits at the center of the semicircle.
A) What is the electric field at the center of the semicircle?
A) The electric vector due to each charge is shown:
Since the distance of each charge to field point is the
same, the magnitude of all E-fields are equal, and let it be E0.
Where [pic]
( E0 = 9.0 ( 102 N/C
From the diagram:
Etotal = E1 + E2 + E3 + E4 + E5
Etotal = –E0 j + E0 cos45 i – E0 cos45 j + E0 i + E0 cos45 i + E0 cos45 j + E0 j
[pic]
A) [pic]
B) What force does the electron experience?
B) [pic]
22. An insulating sphere of radius [pic] has a volume charge density[pic], where a and b are constants.
A) What is the total charge contained in the volume of sphere?
[pic]
[pic]
B) What is the electric field at a distance of r from the center of the sphere?
A spherical Gaussian surface of radius r is shown inside the charged sphere:
[pic]
Electric field everywhere on the surface of the radius r is uniform (not a function of the area) so:
[pic]
[pic]
C) What is the electric field at a distance of 2R from the center of the sphere.
Repeat part A, except that the Gaussian surface is of radius 2R and encloses all of the charge Q of the sphere:
[pic]
[pic] ( [pic]
D) At what radius does the electric field vanish?
From part B: [pic]
Therefore E(r) = 0 at r = 0 or r = 5/4b
23. Two curved plastic rods with uniform linear
charge densities form a semicircle as shown.
One rod carries a total charge +Q and the
other –Q uniformly distributed over their lengths.
Find the magnitude and direction of the electric
field at point P, the center of the semicircle.
The electric fields of an infinitesimal charge – dq and infinitesimal charge +dq are shown on the diagram. The sum these two infinitesimal fields cancel their x-component. Therefore by symmetry the net electric field due to both rods at point P will be only in the y-direction.
[pic]
24. (Bonus Problem) An electron is initially traveling at 3.0 x 106 m/s upward when it enters a uniform electric field of 200 N/C directed to the left. What is the velocity of the electron after it has traveled to a point that has a vertical coordinate that is 10.0 cm upward of where the electron entered the field?
[pic]
As the electron enters the electric field it will accelerate in the opposite direction of the field (+x direction), but will continue its y direction at its original constant speed:
Since the field is constant, so will be the acceleration of the electron, hence we can use kinematics eqs:
y = yo + voy t
0.10m = 0 + 3.0 ( 106 m/s t ( t = 3.3 ( 10–8sec.
So it will take the electron 3.3 ( 10–8sec to reach the point (x , 0.10) in the field.
vx = vox t + ½ ax t2
vx = 0 + ½ (eE/m)t2
vx = ½ (1.6 (10–19C ( 200 N/C /9.11 ( 10-31kg) (3.3 ( 10–8sec)2
vx = 0.020 m/s
Compared to y-component of the velocity this x – component is very small and insignificant. Hence velocity of the electron is at the position of (x , 0.10m) is:
v = 0.020 m/s i + 3.0 ( 106 m/s
magnitude of v is about 3.0 ( 106 m/s and is moving still about y direction.
-----------------------
– – – – – – – – – – – – – – – –
+++++++++++++++++++++
y
E (x,10)
0 x
r
2R
1
2 E5 E4
E4cos 45
3 E2cos 45 E3
E1 E2
4
5
+q
E
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