HW1solutions-10 - Physics 2213 HW #1 Solutions Spring 2010...

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Unformatted text preview: Physics 2213 HW #1 Solutions Spring 2010 0 2 4 q1=? q3=+5 q2=-3 #Q21.10 [Metal Objects] Heres one way to do it. (i) Put the objects together. (ii) Bring a charged object near them. This polarizes the touching objects, inducing negative charge on the left, and positive charge on the right. (iii) Now separate the objects. (iv) And remove the external charge. By charge conservation, the charges on the two metal objects must be equal and opposite. 21.13. We solve this problem by applying Coulombs law. The two forces on 3 q must have equal magnitudes and opposite directions. The force 2 F that 2 q exerts on 3 q has magnitude 2 3 2 2 2 q q F k r = and is in the + x direction. 1 F must be in the x direction, so 1 q must be positive. 1 2 F F = gives 1 3 2 3 2 2 1 2 q q q q k k r r = . 2 2 1 1 2 2 2.00 cm (3.00 nC) 0.750 nC 4.00 cm r q q r = = = . The result for the magnitude of 1 q doesnt depend on the magnitude of 3 q . 21.72. Apply 2 qq F k r = for each pair of charges and find the vector sum of the forces that 1 q and 2 q exert on 3 . q The three charges and the forces on 3 q are shown to the right. (a) 9 9 1 3 9 2 2 4 1 2 2 1 (5.00 10 C)(6.00 10 C) (8.99 10 N m /C ) 1.079 10 N (0.0500 m) q q F k r = = = 36.9 = . 5 1 1 cos 8.63 10 N x F F = = . 5 1 1 sin 6.48 10 N y F F = + = . 9 9 2 3 9 2 2 4 2 2 2 2 (2.00 10 C)(6.00 10 C) (8.99 10 N m /C ) 1.20 10 N (0.0300 m) q q F k r = = = 2 x F = , 4 2 2 1.20 10 N y F F = = . 5 1 2 8.63 10 N x x x F F F = + = . 5 4 5 1 2 6.48 10 N ( 1.20 10 N) 5.52 10 N y y y F F F = + = + = . - 2 - (b) 2 2 4 1.02 10 N x y F F F = + = . tan 0.640 y x F F = = . 32.6 = , below the x + axis. 21.43. [Positive Pair] We find the total electric field by calculating the field due to each charge and taking the vector sum of these two fields. Only the x component of each field is nonzero on the x-axis. The electric field points away from positive charges. (a) Halfway between the two charges, 0. E = (b) For | | x a < , 2 2 2 2 2 4 ( ) ( ) ( ) x q q ax E k kq a x a x x a = = + . For x a > , 2 2 2 2 2 2 2 2 ( ) ( ) ( ) x q q x a E k kq a x a x x a + = + = + . For x a < , 2 2 2 2 2 2 2 2 ( ) ( ) ( ) x q q x a E k kq a x a x x a + = + = + . The magnitude of the field approaches infinity at the location of either one of the point charges. (c) By symmetry, the net electric field is in the +y-direction for y > 0 and in the -y-direction for y < 0....
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This note was uploaded on 06/08/2010 for the course PHYS 213 taught by Professor Perelstein during the Spring '08 term at Cornell University (Engineering School).

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HW1solutions-10 - Physics 2213 HW #1 Solutions Spring 2010...

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