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Wcharges_solutions

Wcharges_solutions - Solutions Set 8 Point Charges and...

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ECE 303 - Sum2014 Solutions - Set 8: Point Charges and Fields 1 8-1. Assume the z coordinate is zero. A 50 nC charge is located at the origin in free space. At any point the E -field has the form | E | u E where u E is the unit vector in the direction of the E -field. Compute | E | and u E at the points below. Express u E in terms of u x and u y . (a) ( x,y ) = (5 , 3), (b) ( x,y ) = ( - 3 , 1) DETAILED SOLUTION: (a) Compute | E | and u E at ( x,y ) = (5 , 3) . The first thing to do is draw a picture which shows the relation between the location of the charge Q and the point ( x,y ) where the field is to be computed. Since the E -field will point radially away from the charge at this point, you can often get an accurate picture of the direction of without doing any math. For example, the figure below shows the geometry for part (a): E r x = 5 y = 3 Q You can see from the figure that the E x and E y components of the E -field and positive, and this should be reflected in your answer for u E . Since u E is in the direction of u r , the unit vector in the radial direction, we can immediately compute u E : u E = u r = r | r | ( a ) The vector r is the vector which starts at the origin (the location of the charge) and ends at ( x,y ) = (5 , 3) (the point where we want to find the field). From the figure this vector is seen to be r = 5 u x + 3 u y | r | = radicalbig 5 2 + 3 2 = 5 . 83 ( b, c ) Substituting ( b ) and ( c ) into ( a ) then gives the following for u E : u E = 5 u x + 3 u y 5 . 83 = 0 . 858 u x + 0 . 515 u y ( d ) In free-space, the magnitude | E | of the E -field is given by equation (4b) in the Notes, repeated here: | E | = 1 4 πǫ 0 Q | r | 2 ( e ) Using Notes equation (3), equation ( c ) above, and Q from the problem statement in ( e ) then gives | E | = 9(10) 9 × 50(10) 9 34 = = 13.24 ( f )
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ECE 303 - Sum2014 Solutions - Set 8: Point Charges and Fields 2 Prob. 8-1 (cont.) (b) Compute | E | and u E at ( x,y ) = ( - 3 , 1) . The figure below shows the geometry for part (b): Q r E x=-3 y=1 Use equation ( a ) and the figure above to compute r as r = - 3 u x + u y | r | = radicalBig ( - 3) 2 + (1) 2 = 3 . 16 ( g,h ) Substituting ( g ) and ( h ) into ( a
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