- #1

- 138

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Here's the problem:

For the arrangement descripbed in the previous problem (see attachment), calculate the electric potential at point B that lies on the perpendicular bisector of the rod a distance b above the x axis.

[lamb] = [alpha] x where [alpha] is a constant.

The correct answer is

V = -(k[alpha]L/2)ln(([squ] [(L^2/4) + b^2)] - L/2)/[squ] [(L^2/4) + b^2)] - L/2))

I'm not getting the same answer.

So far I've got the following: [the] = [<]'a' on the diagram.

V = kq/r

x' = btan[the]

dx' = bsec^2[the]d[the]

x = L/2 + x'

r = bsec[the]

dq = [lamb]d(L/2 + x') = [lamb]dx'

dq = [alpha](L/2 + x')dx' = [alpha](L/2 + btan[the])bsec^2[the]d[the]

so dV = k[alpha](L/2 + btan[the])bsec^2[the]d[the]/bsec[the]

Taking the integral of both sides from -[the] to +[the] doesn't yield the correct result.

I'd appreciate it if someone could point out where I went wrong. I have a feeling the problem's in [lamb] = dq/dL = dq(L/2 + x').

Thanks.