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Calculate the Electric Potential 
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#1
Nov403, 02:57 PM

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. 


#2
Nov403, 03:06 PM

P: 143

I forgot the attachment. Here it is.



#3
Nov403, 03:52 PM

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. 


#4
Nov403, 05:45 PM

P: 143

Calculate the Electric Potential
The rod is not uniformly charged. [lamb] varies with x.



#5
Nov503, 02:25 AM

P: 1,047

Small typo in the answer. Should be:
(kαL/2)ln{( sqrt[(L^{2}/4) + b^{2})] + L/2)/ (sqrt[(L^{2}/4) + b^{2}]  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(b^{2} + (x  L/2)^{2}) so your integral is V = kα∫ (xdx/(sqrt(b^{2} + (x  L/2)^{2})) from x=0 to x=L It actually does work out. Happy integrating. 


#6
Nov503, 09:19 AM

P: 143

Thanks again gnome.



#7
Nov503, 11:42 AM

P: 143

That integral is a royal pain!!!
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? 


#8
Nov503, 04:04 PM

P: 1,047

Actually, nothing. Shoulda done it your way.
dV = kα(L/2 + btanθ)bsec^2θdθ/bsecθ dV = kα(L/2 + btanθ)secθdθ V = kαL/2 ∫secθdθ + kαb ∫tanθsecθdθ V = kαL/2 {lnsecθ + 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[L^{2} + 4b^{2}])/(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[L^{2} + 4b^{2}])/(2b) + L/(2b)}  ln{(sqrt[L^{2} + 4b^{2}])/(2b)  L/(2b)}) Play around with that a little & you get exactly the same answer. 


#9
Nov503, 06:33 PM

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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