@ Abhishekdas
Wow..both of our answers were right. The computer is the only thing that's not doing its job :wink:. For the answer, instead of 1.83*10^-1 Nm^2/C or 1.831*10^-1 Nm^2/C, I used 0.183 Nm^2/C and the computer accepted it.
Thanks guys
EDIT: Just to make sure I understand...
k I got it. So if I were to get the flux of the right side of the cube it would be independent from x and z? and if I took the flux in the front of the cube it would be independent from y and z? or would it just be equal to 0 since E(x,y,z) = Kz j + Ky k?
Oh. I think I know why my integral expression is wrong, it's because I forgot to put dA, dA = k dx dy. And I know that the only component that contributes the flux of the top surface is z.
I came with a second answer of 0.0403 m
x(2) = -q1d/(-q1-q2)
but the computer says I'm giving it the wrong answer. Is there something that I'm doing wrong?
This is what I did but I still get the wrong answers...
a)\int EA = \int (ka)(a^2) = k*a^3/3 = 1.22*10^-1 Nm^2/C
b)\int EA = \int(ka)(a^2) = k*a^3/2 = 1.83*10^-1 Nm^2/C
What am I missing here?
Homework Statement
A 2.96 uC and a -1.85 uC charge are placed 4.48 cm apart. At what points along the line joining them is the potential zero? Let V = 0 at r = infinity and enter the smaller distance first.
Homework Equations
V = k*q/r
The Attempt at a Solution
V(total) = 0 =...
E(top) = Kz j
Is this right? I tried substituting the values for that, and I tried something different for getting the integral of E(top) but still got the wrong answers for both tries.