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PH106 Practice Exam I Fall , 2009
- 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:
๐ธ 1 + ๐ธ 2 = 0 + ๐
cos ๐ = 9 โ 109 ๐ โ ๐^2 /๐ถ^2
(. 3^2 +. 2^2 )๐^2
ร
. 3^2 +. 2^2
= 1.2 ร 10^5 ๐/๐ถ
WRT axis Y:
๐ธ 1 + ๐ธ 2 = ๐
sin ๐
= 9 โ 109 ๐ โ ๐^2 /๐ถ^2
. 3^2 ๐^2
. 3^2 +. 2^2 ๐^2
ร
. 3^2 +. 2^2
= 2.3 ร 10^5 ๐/๐ถ
Angle is
tanโ^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 electronโs 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: 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
= 2 ร .15๐ ร
1.6 โ 10 โ^19 ๐ถ ร 200๐/๐ถ
9.11 โ 10 โ^31 ๐๐
= 3.2 ร 10^6 ๐/๐
- 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.
๐incl ๐ 0
8.85 โ 10 โ^12 ๐ถ^2 /๐ โ ๐
= 2.3 ร 10^11 ๐ โ ๐/๐ถ
- 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 = 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 =
2.3 ร 10^5 ๐๐/๐ถ
2 ร 3.14 ร .05๐
= 7.2 ร 10^5 ๐/๐ถ
- 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
