P206 E1spr 2011 Feb 14, 2011
SOLUTIONS
PHYS 206 C UNIVERSITY PHYSICS II FEB. 14,
2011
Dr. S.F. Ahmad EXAM I MM=30
Time: 50 min.
NOTE: Please answer all the questions. Show your working or explain your answer for partial
credits.
Multiple choice questions:(work is not required to be shown)
M.1[1].The electric flux through a surface of fixed area is a maximum when the surface is
a. parallel to the electric field. b. antiparallel to the electric field.
c. perpendicular to the electric field. d. at an angle of π / 4 radians to the electric field.
e. closed, but does not contain the charge.
M.2 [1]. The potential energy of a pair of unlike charges is
a. positive. b. negative. c. neutral. d. proportional to the square of the distance.
e. inversely proportional to the square of the distance.
M.3[3] Given the electric potential function V = 2xy -3xz + 5y2. Calculate the x- component of the electric field at electric
field at x = 1 m, y = 1 m, z = 1 m.
Soln: Differentiate the given equation for v with respect to x and get
E = - dV/dx = -2y + 3z; for the given value of x=y=1 we get electric field as E = 1 V/m
Ans: E = 1.0 V/m
You must show the equations used and your work for Q1 to Q5:
Q.1 [5] A long thin rod has a charge density λ = 3.0 µC/m.C/m. Find the electric field strength 1.0 m from the center of
the rod measured perpendicular to the rod’s symmetrical axis.
Let the rod be in the x direction then along the y-direction the electric field is
Soln: Ey = 2 k λ / y = 2 * (8.99x109 Nm2/C2) * (3.0x10-6 C/m) / 1.0 m
Ans: 5.39x104 N/C
With
Ey = k λ / y = ½ (5.39x104) = 2.70x104 N/C this answer will also be taken as correct because
wrong equation was written on the blackboard.
Q.2 [5] A charge q= -40 mc is located on the x-axis at x = 0, and another charge Q= 30 mC is placed at x = 2.0 m.
Determine the net electric flux Ф, in Nm2/C, through a spherical surface of radius = 1.5 m centered on the origin
that encloses charge q.
Soln: ФE = qin / o we see that the only charge that is inside the spherical surface of radius 1.5 m is q = -40 nC,
hence
ФE = ( - 40 x 10-9 C ) / (8.85 x 10-12 C2 / N . m2) = - 4.52 x 103 Nm2/C
Ans: - 4.52 x 103 Nm2/C
