No Title
PHYSICS II
Solution to Exam 4
                                                                                                           
Friday 2/12/16 Name
O233: periods 1,2
JFW CM:
Period:














                    
                                         (1)
(25)
(2)
(25)
(3)
(25)
(4)
(25)
TOTAL
(100)
  1. The figure shows a surface that is completely closed. There are two electric fields producing electric flux. One is a constant field, the other varies with position.
    E(x)
    =
    [(450 N/C·m)x+50.0 N/C] 
    ^
    i
     
    E(y)
    =
    800 N/C 
    ^
    j
     

    a=12.0 cm
    b=10.0 cm
    c=8.00 cm

      (a) [5 pt s] Calculate the net flux through the enclosure due to E(y).
      ANS: There is no net electric flux for this electric field because it is a uniform electric field. The flux that enters the enclosure equals the flux that leaves the enclosure.
      (b) [10 pts] Calculate the net flux through the enclosure due to E(x).
      Front Face:
      FF
      =
      E(a)bc
      =
      [(450 N/C·m)(0.120 m)+50.0 N/C][ 1
      2
      		 (0.100 m)(0.0800 m)]
      =
      0.416 N·m2/C
      Back Face:
      FB
      =
      -E(0)bc
      =
      -[(450 N/C·m)(0)+50.0 N/C][ 1
      2
      		 (0.100 m)(0.0800 m)]
      =
      -0.200 N·m2/C
      The net electric flux through the enclosure is
      Fnet
      =
      (0 + 0.416 - 0.200) N·m2/C
      =
      0.216 N·m2/C

      (c) [10 pts] Calculate the net charge inside the enclosure. The net charge inside is
      qnet
      =
      eoFnet
      =
      (8.85×10-12 C2/N·m2) (0.216 N·m2/C)
      =
      +1.91 pC
  2. The figure shows a positively charged metal sphere of radius a with static charge, Q. There is a concentric shell surrounding the sphere with inner radius b and outer radius c containing static excess charge, qexc. There are four field points: A, B, C, and D.
    a=5.00 cm
    b=7.00 cm
    c=8.00 cm
    Q=+8.20 mC
    qexc=-12.0 mC

    rA=3.00 cm
    rB=6.00 cm
    rC=7.50 cm
    rD=10.0 cm

      (a) [5 pts] How much charge is on the inner surface of the shell?
      From Gauss's Law
      qin=-Q=-8.20 mC

      (b) [4 pts] How much charge is on the outer surface of the shell?

      qexc
      =
      qin+qout
      qout
      =
      qexc-qin
      =
      (-12.0 mC)-(-8.20 mC)
      =
      -3.80 mC

      (c) [16 pts] Determine the electric fields at points A, B, C, and D.
      EA
      =
      0
      EC
      =
      0
      EB
      =
      kQ
      r2B
      =
      (8.99×109 N·m2/C2) (+8.20×10-6 C)
      (0.0600 m)2
      =
      +2.05×107 N/C
      ED
      =
      kqout
      r2D
      �
      kqnet
      r2D
      =
      (8.99×109 N·m2/C2) (-3.80×10-6 C)
      (0.100 m)2
      =
      -3.42×106 N/C
  3. Two identical, positive point charges are located on the y-axis positioned symetrically about the origin.
    q=5.00 mC
    a=30.0 cm
    The electric potential at an arbitrary point on the x-axis is
    V(x)
    =
    2kq

    �
    a2+x2

      (a) [10 pts] Calculate the potential difference VAB between the points xA=0.800 m and xB=1.20 m.

      VAB
      =
      VB-VA
      =
      2kq �
      �
      �       		1

      �
      a2+x2B
      		 - 1

      �
      a2+x2A
      		 �
      �
      �
      =
      2(8.99×109 N·m2/C2) (5.00×10-6 C)
      �
      �
      �       		1

      �
      (0.300 m)2+(1.20 m)2
      		 - 1

      �
      (0.300 m)2+(0.800 m)2
      		 �
      �
      �
      =
      -3.25×104 V

      (b) [10 pts] Use the derivative method for V(x) to find the expression for the electric field, E(x) for any point on the x-axis. DO NOT USE COULOMB'S FIELD EQUATION.

      V(x)
      =
      2kq(a2+x2)-1/2
      E(x)
      =
      - �V
      �x
      =
      -2kq �
      �       		- 1
      2
      		 (a2+x2)-3/2 (2x) �
      �
      =
      2kqx
      (a2+x2)3/2

      (c) [5 pts] Does the electric field have a maximum value in the range 0 � x � �? Explain your answer.
      ANS: Yes, there is a maximum. There is a zero field at x=0 and a zero field at x=�. Since the electric field is non-zero between these to limits, the field must go through a maximum. This can be shown by either plotting the function or setting the derivative of the electric field equal to zero and solving for the point at which the field is a maximum. The maximum occurs at x=a/�2.
  4. The figure shows a sphere made of insulated material. It is uniformly charged with positve charge (r = constant). The electric potential inside the sphere is
    Vin(r)
    =
    3kQ
    2R
    		 - �
    �     		kQ
    2R3
    		 �
    �     		r2

    R=12.0 cm
    Q=+8.00 mC

      (a) [5 pts] What charge within the sphere contributes to the electric potential at the field point P?
      ANS: ALL of the charge in the sphere contributes to the electric potential at the field point, P.
      (b) [6 pts] Calculate the maximum electric potential in the system (0 � r � �).
      The maximum electric potential occurs at r=0.
      V(0)
      =
      3kQ
      2R
      =
      3(8.99×109 V·m/C) (+8.00×10-6 C)
      2(0.120 m)
      =
      8.99×105 V

      (c) [7 pts] Calculate the electric potential at r=15.0 cm.

      Vout
      =
      kQ
      r
      =
      (8.99×109 V·m/C) (8.00×10-6 C)
      0.150 m
      =
      +4.79×105 V

      (d) [7 pts] From Vin(r), use the derivative method to find Ein(r).

      Ein(r)
      =
      - �Vin(r)
      �r
      =
      - �
      �       		0-2 �
      �       		kQ
      2R3
      		 �
      �       		r �
      �
      =
      �
      �       		kQ
      R3
      		 �
      �       		r
CONSTANTS

m=10-3
m = 10-6
k=8.99×109 N·m2/C2
eo=8.85×10-12 C2/N·m2
EQUATIONS

U=kq1q2/r
V=kq/r
W=q(V2-V1)
Vb-Va=- �
� 		b

a 
 E·dl
FE=�
(�)
�
E·
^
n
 
 dA
FE=qin/ eo
Ex=- �V
�x
Ey=- �V
�y
Ez=- �V
�z
E=kq/r2
E=-�V/�r


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On 17 Feb 2016, 10:38.