Question
A +33.0 nC charge is placed at the origin of an xy-coordinate system, and a −25.0 nC charge is placed at the point (1.500 m, −0.0540 m).
(a) Determine the magnitude and the direction of the force that the positive charge exerts on the negative charge. Use ijk notation for your result.
ANS: -(3.29e-6)i +(1.18e-7)j N
That is the right answer, however I don't get the how they got the y position> i get 2.55e-3 .
Im not sure if i'm missing something, please help me out!
Thanks
(a) Determine the magnitude and the direction of the force that the positive charge exerts on the negative charge. Use ijk notation for your result.
ANS: -(3.29e-6)i +(1.18e-7)j N
That is the right answer, however I don't get the how they got the y position> i get 2.55e-3 .
Im not sure if i'm missing something, please help me out!
Thanks
Answers
bobpursley
F=l q1q2( r2-r1)/(r2-r1)^3
so , if the i is right, lets look only at the j component
Fj=kq1Q2(r2j-r1j)/( )^3
but r1 is the origin, so
r2=1.5i-.054j)
so the distance in the i is 1.5/.054= 27,7 times the j distance, so the force then should be 27.7 times less in that direction vector.
so, however you got the 3.29e-6, divide it by 27.7 to get 1.18e-7.
So this indicates you made a math error in computing the j direction only, probably calc error, all the other computes on the basic formula fraction are correct.
I don't quite understand why we had to do 1.5/.054 ? Like what is the reasoning behind it?
Why didn't we just do
F=(q1q2)/r^2
where r would now be 0.054? and if we can't do this why is it wrong to do this?
Really appreciate the help!
Why didn't we just do
F=(q1q2)/r^2
where r would now be 0.054? and if we can't do this why is it wrong to do this?
Really appreciate the help!
NO!
r = sqrt (1.5^2 + .054^2)
r^2 = 2.25+ .003 = 2.253
|F| = k q1 q2 /r^2
Fi = |F| cos angle above -xaxis
Fj = |F| sin angle above -x axis
that angle is tiny tan angle = .054/1.5
so cos angle is = 1
and sin angle = tan angle = .054/1.5
so
Fj = |F| * .054/1.5 = 1/27.7
as Bob Pursley was saying.
r = sqrt (1.5^2 + .054^2)
r^2 = 2.25+ .003 = 2.253
|F| = k q1 q2 /r^2
Fi = |F| cos angle above -xaxis
Fj = |F| sin angle above -x axis
that angle is tiny tan angle = .054/1.5
so cos angle is = 1
and sin angle = tan angle = .054/1.5
so
Fj = |F| * .054/1.5 = 1/27.7
as Bob Pursley was saying.