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Material Type: Exam; Professor: Mryasov; Class: Generl Physics W/Calc II; Subject: PH-Physics; University: University of Alabama; Term: Fall 2009;
Typology: Exams
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PH106 Practice Exam I Fall , 2009
Solution:
WRT axis X:
๐ธ 1 + ๐ธ 2 = 0 + ๐
cos ๐ = 9 โ 109 ๐ โ ๐^2 /๐ถ^2
WRT axis Y:
๐ธ 1 + ๐ธ 2 = ๐
sin ๐
Angle is
tanโ^1
P
q 1 = 4 ๏ญC q 2 = - 3 ๏ญC
x
y
0.30 m
0.2 m
Solution: ๐๐ฅ = ๐๐๐๐ ๐ก = ๐ 0 = 3 ร 10^6 ๐/๐ One can find the y-component of velocity from the equation: ๐ =
๐๐^2 โ๐๐^2 2 ๐ , and^ ๐๐^ =^ ๐๐ธ
๐incl ๐ 0
(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 = ๐incl ๐ 0 =^
๐๐ ๐ 0 =^
๐/๐ฟโ๐ ๐ 0 =
2 ๐ ๐ถ .5๐ โ.1๐ 8.85ร10โ^12 ๐ถ^2 /๐๐ = 2.3 ร 10
(b) What is the approximate value of the electric field on the outer surface of the cylinder?
ฮฆ = 2๐๐๐ธ โ ๐ธ =
b
a