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Exam 1 Solution - Physics II Intermediate - 2011 |, Exams of Physics

Material Type: Exam; Class: Physics 2 - Intermediate; Subject: Physics; University: Bellarmine University; Term: Forever 1989;

Typology: Exams

2010/2011

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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
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P206 E1spr 2011 Feb 14, 2011

SOLUTIONS

PHYS 206 C UNIVERSITY PHYSICS II FEB. 14,

D r. S.F. Ahmad EXAM I MM=

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 + 5y^2. 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 * (8.99x10^9 Nm^2 /C^2 ) * (3.0x10-6^ C/m) / 1.0 m

Ans: 5.39x10^4 N/C

With

Ey = k λ / y = ½ (5.39x10^4 ) = 2.70x10^4 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 Nm^2 /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^ C^2 / N. m^2 ) = - 4.52 x 10^3 Nm^2 /C

Ans: - 4.52 x 10^3 Nm^2 /C

P206 E1spr 2011 Feb 14, 2011 Q.3 [5] Two point charges are arranged such that q 1 = 6.00 nC is at (0.3 m, 0), q 2 = -3.00 nC is at (0, -0.1 m)

according to a rectangular coordinate system. Calculate the magnitude of the electric potential at a point P located

at the origin.

Soln: Use VP = k (q 1 /r + q 2 /r) = 8.99x10^9 (6x10-9/(.3) + (-3.0x10-9) / (0.1)) = -89.9 V

Ans: -89.9 V

Q.4 [5] Points A and B lie 20 cm apart on a line extending radially

from a point charge Q, and the potentials at these points are VA= 280 V

and VB = 130 V. Find Q and the distance, r, between A and the charge.

VA = kQ/r = 280 or kQ = 280 r ……(1) and VB = kQ /(0.2 + r) = 130 or kQ = 130 (0.2 + r) …..(2) From (1) and (2) 280 r = 130 (0.2 + r) get r = 0.173 m and Q = 5.39x10-9^ C Ans: r = 0.173 m and Q = 5.39x10-9^ C Q.5[5] How strong an electric field is needed to accelerate electrons in a TV tube from rest to one-tenth the speed

of light in a distance of 5.0 cm?

Soln: Method I: Δk = kk = kf – ki = ½ me * vf^2 – 0 = ½ * 9.1x10-31^ * (0.13x10^8 )^2 = 4.095x10-16^ ; ki , kinetic energy is zero initially. Potential energy change Δk = kU = - Δk = kk; and Δk = kU = q Δk = kV = - q* d* E ; d = distance travelled, E is the electric field Hence E = Δk = kk / (q* d) Calculate and get electric field , E = 5.1x10^4 V/m as the answer. Ans: E = 5.1x10^4 0.3 m q 1 -0.1m q X Y P Q A B r