PHYS1322-CH21Lec2

# PHYS1322-CH21Lec2 - Lecture 3 Lecture 3 Lecture 3 Last week...

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Lecture 3 Lecture 3 Lecture 3 Superposition Last week: Coulomb’s Law 10 10 10 2 01 0 1 ˆ 4 qq F r r πε = G 10 20 ....... FF F =++ G GG q 0 q 1 10 F G 10 ˆ r 10 r q 2 20 F G F G An Example What is the force acting on q o () ? q o , q 1 , and q 2 are point charges q o = -1 µ C at (1,3)cm, q 1 = 3 µ C at (1,0)cm, q 2 = 4 µ C at (5,0)cm. 0 F G Decompose into its x and y components 20 ˆ r 20 02 ˆˆ ˆ cos sin 43 = - 55 rx y xx yy x y rr xy θ =+ −− + 10 2 10 kq q F r = 10 2 10 ˆ F k y r = G 20 2 20 kq q F r = 20 20 2 20 ˆ F kr r = G We have 20 10 F F G G and Find 0 F FF GGG 10 10 10 ˆ F Fr = G x ( cm ) y ( cm ) 1 2 3 4 5 4 3 2 1 q o q 2 q 1 F 20 F 10 An Example continued q o , q 1 , and q 2 are point charges q o = -1 µ C at (1,3)cm, q 1 = 3 µ C at (1,0)cm, q 2 = 4 µ C at (5,0)cm. x ( cm ) y ( cm ) 1 2 3 4 5 4 3 2 1 q o q 2 q 1 10 F G 20 F G 0 F G Let’s put in the numbers . . . 10 .03m r = 22 20 .03 .04 .05m r = 92 2 =9 10 kN m C 10 2 66 9 4 ˆ 110 310 ˆ 9 10 910 ˆ 30 Fk y r y yN = −⋅ ⋅⋅ =⋅ =− G 96 6 20 4 20 410 25 10 14.4 14.4 (11.5 8.6 ) N F N Fx y ⋅ ⋅ =  +   G Now add the components of and to find An Example continued 0 F G q o , q 1 , and q 2 are point charges q o = -1 µ C at (1,3)cm, q 1 = 3 µ C at (1,0)cm, q 2 = 4 µ C at (5,0)cm.

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## This note was uploaded on 04/14/2008 for the course PHYS 1322 taught by Professor Ndili during the Fall '07 term at University of Houston.

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PHYS1322-CH21Lec2 - Lecture 3 Lecture 3 Lecture 3 Last week...

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