# Calculate the Electric Potential

by discoverer02
Tags: electric, potential
 P: 143 What am I doing wrong here? [*(][*(][*(][*(] 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.
P: 143
I forgot the attachment. Here it is.
Attached Images
 diagram.bmp (23.7 KB, 36 views)
 P: 513 How is that rod charged? You say dq = λ dx' with λ = α x. This would mean that the rod is NOT uniformly charged. Are you sure this is what the problem says? Normally, problems like this have a uniformly charged rod, meaning sth. like dq = λ dx', where λ is a constant.
P: 143

## Calculate the Electric Potential

The rod is not uniformly charged. [lamb] varies with x.
 P: 1,048 Small typo in the answer. Should be: -(kαL/2)ln{( sqrt[(L2/4) + b2)] + L/2)/ (sqrt[(L2/4) + b2] - L/2)} (the answer you posted equals 0) Don't set it up in terms of the angle. You want V = -k∫dq/r dq = Λdx and Λ = αx therefore: dq = αxdx and r = sqrt(b2 + (x - L/2)2) so your integral is V = -kα∫ (xdx/(sqrt(b2 + (x - L/2)2)) from x=0 to x=L It actually does work out. Happy integrating. [:)]
 P: 143 Thanks again gnome.
 P: 143 That integral is a royal pain!!! [o)] I'm curious. What was wrong with the integral I was originally working with other than the fact that it might be easier integrating with respect to 'x' rather than the angle?
 P: 1,048 Actually, nothing. Shoulda done it your way. [g)] [g)] [g)] dV = kα(L/2 + btanθ)bsec^2θdθ/bsecθ dV = kα(L/2 + btanθ)secθdθ V = kαL/2 ∫secθdθ + kαb ∫tanθsecθdθ V = kαL/2 {ln|secθ + tanθ|}{from -θ to θ} + kαb{secθ}{from -θ to θ} The way you have defined theta, up at the top there, sec(-θ) = sec(θ) = (sqrt[L2 + 4b2])/(2b) so the second term of the integral drops out completely. tanθ = L/(2b) tan(-θ) = -L/(2b) so the integral, evaluated from -θ to θ, becomes: (kαL/2)(ln{(sqrt[L2 + 4b2])/(2b) + L/(2b)} - ln{(sqrt[L2 + 4b2])/(2b) - L/(2b)}) Play around with that a little & you get exactly the same answer.
 P: 143 Thanks. It's a relief to know I set the integral up correctly. A stupid mistake on my part evaluating the integral turned into something it's not, so I thought I was going about it the wrong way.

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