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General Physics with Calculus II - Practice Exam 1 | PH 106, Exams of Physics

Material Type: Exam; Professor: Mryasov; Class: Generl Physics W/Calc II; Subject: PH-Physics; University: University of Alabama; Term: Fall 2009;

Typology: Exams

Pre 2010

Uploaded on 11/24/2009

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PH106 Practice Exam I Fall , 2009
1. Find the magnitude and direction with respect to
the x-axis of the electric field at point P in the
diagram to the right.
Solution:
WRT axis X:
๐ธ1
+๐ธ2
= 0 + ๐‘˜๐‘ž2
๐‘Ÿ2
2cos ๐œƒ= 9 โˆ™109๐‘โˆ™๐‘š2/๐ถ23๐œ‡๐ถ
(. 32+. 22)๐‘š2ร—. 2
. 32+. 22
= 1.2 ร— 105๐‘/๐ถ
WRT axis Y:
๐ธ1
+๐ธ2
=๐‘˜๐‘ž1
๐‘Ÿ1
2โˆ’๐‘˜๐‘ž2
๐‘Ÿ2
2sin ๐œƒ
= 9 โˆ™109๐‘โˆ™๐‘š2/๐ถ2 4๐œ‡๐ถ
. 32 ๐‘š2โˆ’3๐œ‡๐ถ
. 32+. 22 ๐‘š2ร—. 3
. 32+. 22
= 2.3 ร— 105๐‘/๐ถ
Angle is
tanโˆ’1๐ธ๐‘ฆ
๐ธ๐‘ฅ=62ยฐ
2. An electron is projected horizontally with a speed v0 = 3 x 106 m/s into a region
where there is a uniform electric field E = 200 N/C directed down. What are the x-
and y- components of the electronโ€™s velocity after it has traveled 0.15 m in the x-
direction?
P
q2 = - 3 ๏ญC
q1 = 4 ๏ญC
x
y
0.30 m
0.2 m
pf3

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PH106 Practice Exam I Fall , 2009

  1. Find the magnitude and direction with respect to the x-axis of the electric field at point P in the diagram to the right.

Solution:

WRT axis X:

๐ธ 1 + ๐ธ 2 = 0 + ๐‘˜

cos ๐œƒ = 9 โˆ™ 109 ๐‘ โˆ™ ๐‘š^2 /๐ถ^2

(. 3^2 +. 2^2 )๐‘š^2

ร—

. 3^2 +. 2^2

= 1.2 ร— 10^5 ๐‘/๐ถ

WRT axis Y:

๐ธ 1 + ๐ธ 2 = ๐‘˜

sin ๐œƒ

= 9 โˆ™ 109 ๐‘ โˆ™ ๐‘š^2 /๐ถ^2

. 3^2 ๐‘š^2

. 3^2 +. 2^2 ๐‘š^2

ร—

. 3^2 +. 2^2

= 2.3 ร— 10^5 ๐‘/๐ถ

Angle is

tanโˆ’^1

  1. An electron is projected horizontally with a speed v 0 = 3 x 10^6 m/s into a region where there is a uniform electric field E = 200 N/C directed down. What are the x- and y- components of the electronโ€™s velocity after it has traveled 0.15 m in the x- direction?

P

q 1 = 4 ๏ญC q 2 = - 3 ๏ญC

x

y

0.30 m

0.2 m

Solution: ๐‘‰๐‘ฅ = ๐‘๐‘œ๐‘›๐‘ ๐‘ก = ๐‘‰ 0 = 3 ร— 10^6 ๐‘š/๐‘  One can find the y-component of velocity from the equation: ๐‘‘ =

๐‘‰๐‘“^2 โˆ’๐‘‰๐‘–^2 2 ๐‘Ž , and^ ๐‘š๐‘Ž^ =^ ๐‘’๐ธ

= 2 ร— .15๐‘š ร—

1.6 โˆ™ 10 โˆ’^19 ๐ถ ร— 200๐‘/๐ถ

9.11 โˆ™ 10 โˆ’^31 ๐‘˜๐‘”

= 3.2 ร— 10^6 ๐‘š/๐‘ 

  1. In the figure, the dotted curve represents a closed surface. All charges have the value q = ๏‚ฑ 1.0 ๏ญC. Find the electric flux out of the surface.

๐‘žincl ๐œ– 0

8.85 โˆ™ 10 โˆ’^12 ๐ถ^2 /๐‘ โˆ™ ๐‘š

= 2.3 ร— 10^11 ๐‘ โˆ™ ๐‘š/๐ถ

  1. A rod of length L = 0.5 m has a charge Q = 2 ๏ญC uniformly distributed along the rod.

(a) What is the electric flux through a cylindrical surface of length d = 0.1 m and radius r = 0.05 m near the center of the rod, as shown above? The cylindrical surface encloses a center portion of the rod.

Solution: ฮฆ = ๐‘žincl ๐œ– 0 =^

๐œ†๐‘‘ ๐œ– 0 =^

๐‘„/๐ฟโˆ™๐‘‘ ๐œ– 0 =

2 ๐œ‡ ๐ถ .5๐‘š โˆ™.1๐‘š 8.85ร—10โˆ’^12 ๐ถ^2 /๐‘๐‘š = 2.3 ร— 10

(b) What is the approximate value of the electric field on the outer surface of the cylinder?

ฮฆ = 2๐œ‹๐‘Ÿ๐ธ โ‡’ ๐ธ =

2.3 ร— 10^5 ๐‘๐‘š/๐ถ

2 ร— 3.14 ร— .05๐‘š

= 7.2 ร— 10^5 ๐‘/๐ถ

  1. A point charge 2Q is at the center of a hollow metal sphere of inner radius a and outer radius b. The total charge on the hollow sphere is โ€“ Q.

b

a