A point charge is at the origin. With this point charge as the source point, wha

Question

A point charge is at the origin. With this point charge as the source point, what is the unit vector r in the direction of (a) the field point at x = 0, y = -1.35m; (b) the field point at x = 12.0cm, y = 12.0cm; (c) the field point at x = -1.10m, y = 2.60m? Express your results in terms of the unit vectors i and j. You may want to review (pages 695-699) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Electric field vector for a point charge Express your answer in terms of the unit vectors i and j. Express your answer in terms of the unit vectors i and j. Express your answer in terms of the unit vectors i and j.

Explanation / Answer

let a=(0,0) be the origin and b=(x,y) be the other point
then the vector from a to b is v=b-a = (x,y) - (0,0) =<x,y>
to get the unit vector you divide <x,y> by it's modulus
|x,y| = sqrt( x^2 + y^2)
so u= <x,y> / |x,y|
a) v=<0,-1.35> and |v|= sqrt( 0^2 + (-1.35)^2) =1.35
so u= <0,-1.35> / 1.35
=<0,-1>
= -1j
b) v=<12,12> and |v|= sqrt( 12^2 + 12^2) =12*sqrt2
so u= <12,12> / 12*sqrt2
=<1/sqrt2 , 1/sqrt2>
= 1/sqrt2 i +1/sqrt2 j
c) v=<-1.1,2.6> and |v|= sqrt( 1.1^2 + 2.6^2) =sqrt7.97
so u= <1.1,2.6> / sqrt7.97
=<1.1/sqrt7.97 , 2.6/sqrt7.97>
= 1.1/sqrt7.97 i + 2.6/sqrt7.97 j

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