Physics_51_Prof_Test_1s_key

Physics_51_Prof_Test_1s_key - E x = k (2 •10-9 / 3 2- 3...

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Physics 51 Proficiency Test 1(sample) (Time: 15 minutes) 25 points By Todd Sauke Name ___________ KEY ________ Section # __ KEY ___ The three charges shown below as black dots are the source of an electric field. The coordinates are measured in meters, and the charges are as indicated in units of nano- Coulombs (nC = 10 -9 C). Find the magnitude and direction of the electric field vector at the origin, O . Pay particular attention to the sign of all quantities. Use k = 1/(4 π ε 0 ) = 9.0 x 10 9 N m 2 /C 2 . 2 nC -5 nC 3 nC O -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 5 x y We add the respective x & y components from the three charges, and use the fact that the x component is given by –kq/r 2 times the cosine of the angle the charge makes with the +x axis. The y component = –kq/r 2 times the sine of the angle.
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Unformatted text preview: E x = k (2 •10-9 / 3 2- 3 •10-9 / 4 2 + (5 •10-9 / (2 • 3 2 )) cos(45º) ) (note: cosine is in numerator!) = 9 •10 9 (0.2222 •10-9- 0.1875 •10-9 + 0.2778 •10-9 • 0.707) = 9 (0.231) = 2.080 E y = k (0.0 + 0.0 + (5 •10-9 / (2 • 3 2 )) sin(45º) ) (note: sine is in numerator! = 9 •10 9 (0.0 + 0.0 + 0.2778 •10-9 • 0.707) = 9 (0.1964) = 1.768 (you must check the quadrant of the angle!) Magnitude = sqrt(Ex 2 + Ey 2 ) = sqrt(2.080 2 + 1.768 2 ) = 2.730 θ = tan-1 (Ey / Ex) = tan-1 (1.768 / 2.080) = 40.4º x-component at O = __ 2.080 _____N/C y-component at O = ____ 1.768 ______ N/C Magnitude of E field at O __ 2.730 ___ N/C Direction of E field at O ___ 40.4 ___ º ( r e l a t i v e t o + x a x i s )...
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This note was uploaded on 09/08/2010 for the course PHYS 51 at San Jose State University .

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