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.
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
3 answers
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.
1 answer