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Calculate the Electric Potential

by discoverer02
Tags: electric, potential
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discoverer02
#1
Nov4-03, 02:57 PM
P: 142
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.
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discoverer02
#2
Nov4-03, 03:06 PM
P: 142
I forgot the attachment. Here it is.
Attached Images
File Type: bmp diagram.bmp (23.7 KB, 41 views)
arcnets
#3
Nov4-03, 03:52 PM
P: 513
How is that rod charged?

You say
dq = &lambda; dx' with
&lambda; = &alpha; 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 = &lambda; dx', where &lambda; is a constant.

discoverer02
#4
Nov4-03, 05:45 PM
P: 142
Calculate the Electric Potential

The rod is not uniformly charged. [lamb] varies with x.
gnome
#5
Nov5-03, 02:25 AM
P: 1,046
Small typo in the answer. Should be:

-(k&alpha;L/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&int;dq/r
dq = &Lambda;dx and &Lambda; = &alpha;x
therefore:
dq = &alpha;xdx
and
r = sqrt(b2 + (x - L/2)2)

so your integral is

V = -k&alpha;&int; (xdx/(sqrt(b2 + (x - L/2)2)) from x=0 to x=L

It actually does work out. Happy integrating.
discoverer02
#6
Nov5-03, 09:19 AM
P: 142
Thanks again gnome.
discoverer02
#7
Nov5-03, 11:42 AM
P: 142
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?
gnome
#8
Nov5-03, 04:04 PM
P: 1,046
Actually, nothing. Shoulda done it your way.

dV = k&alpha;(L/2 + btan&theta;)bsec^2&theta;d&theta;/bsec&theta;
dV = k&alpha;(L/2 + btan&theta;)sec&theta;d&theta;
V = k&alpha;L/2 &int;sec&theta;d&theta; + k&alpha;b &int;tan&theta;sec&theta;d&theta;
V = k&alpha;L/2 {ln|sec&theta; + tan&theta;|}{from -&theta; to &theta;} + k&alpha;b{sec&theta;}{from -&theta; to &theta;}
The way you have defined theta, up at the top there, sec(-&theta;) = sec(&theta;) = (sqrt[L2 + 4b2])/(2b) so the second term of the integral drops out completely.

tan&theta; = L/(2b)
tan(-&theta;) = -L/(2b)

so the integral, evaluated from -&theta; to &theta;, becomes:
(k&alpha;L/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.
discoverer02
#9
Nov5-03, 06:33 PM
P: 142
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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