Arturo O. answered • 02/01/20
Experienced Physics Teacher for Physics Tutoring
(a)
Ex(x) = kq1/(x1 - x)2 + kq2/(x2 - x)2
x = 0 ⇒
Ex(0) = k(q1/x12 + q2/x22)
Plug in the values of k, q1, x1, q2, and x2 in appropriate units with correct signs:
k = 9 x 109 Nm2/C2
q1 = 9 x 10-6 C
x1 = 0.032 m
q2 = -24 x 10-6 C
x2 = 0.081 m
You will get Ex(0) in units of N/C.
(b)
Work the same way as part (a), but this time x is 0.057 m.
Ex(x) = kq1/(x1 - x)2 + kq2/(x2 - x)2
Plug in the same values of k, q1, x1, q2, and x2 in appropriate units with correct signs as in part (a), and also plug in
x = 0.057 m
You do the number crunching and finish from here.
Arturo O.
I got 4.62e7 as the strength of the field in part (a), in units of N/C (close to your answer). However, the x component of that field must point to the left of the location of q1, since the field lines radiate out of a positive x1. That makes the x component negative to the left of x1, but positive to the right of x1. Try entering -4.62e7 for part (a) and let me know if it worked. You will probably have to make the same kind of adjustment for part (b).02/02/20
Joseph D.
02/02/20
Arturo O.
But q2 = -24E-6, so the term with q2 is negative. I think you added field strengths from q1 and q2 instead of taking the difference. The source charges at x1 and x2 have opposite signs. If I add the strengths, I get 1.12E8, as you did, but the opposite signs mean we need to take the difference of the field strengths.02/02/20
Cass K.
not too sure what I'm doing wrong. I got a) 4.6e7 and b) -3.7e8 which were both wrong :(02/01/20