Homework Help: Finding Voltage at a point

1. Feb 23, 2014

iScience

I want to find the potential at point P due to a finite, constant linear charge density λ, of length L.

first, if i wanted to find the x-component of the E-field at point P..

$$E=\int{\frac{k(λdx)}{r^2}}cosθ$$

$$r=\sqrt{z^2+x^2}$$

$$cosθ=\frac{x}{\sqrt{z^2+x^2}}$$

$$E=kλ\int{\frac{1}{z^2+x^2}}\frac{x}{\sqrt{z^2+x^2}}dx$$

evaluated from 0 to L yields..

$$-kλ(\frac{1}{z^2+L^2}-\frac{1}{\sqrt{z^2}})$$

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when we did found the potential at piont P in class, we did the following though

$$V_{p}= -\int{\vec{E}\cdot d\vec{r}cosθ}$$

my confusion is, when we add the cosθ term, aren't we just finding the voltage of the x-component E-field?? Because, i thought we were trying to find the total voltage at point P, but this suggests otherwise doesn't it?

also..

expanding $$V_{p}$$..

since $$r=\sqrt{z^2+x^2}$$

$$dr=\frac{1}{2}(z^2+x^2)^{-1/2}(2xdx)$$

$$V_{p}= \int{\frac{k(λdx)}{z^2+x^2}\frac{x}{\sqrt{z^2+x^2}}(\frac{1}{2}(z^2+x^2)^{-1/2}(2xdx))=kλ\int{\frac{x^2}{(z^2+x^2)^2}}dx}$$

(evaluated from 0 to L)

is this expression for the potential correct? in class we got..

$$-kλ\int{\frac{zx}{z^2\sqrt{z^2+x^2}}\frac{x^2}{z^2+x^2}dx}$$

Last edited: Feb 24, 2014
2. Feb 23, 2014

rude man

Why mess with E fields at all? Use dV = k dq/r etc. and integrate.

Your last integral does not make sense ...

3. Feb 24, 2014

iScience

other than that does it make sense? moreover is it the correct expression for the voltage at point P?

using dV=kq/r i get..

$$V=\int{\frac{k}{r}dq}=kλ\int{\frac{1}{\sqrt{x^2+z^2}}dx}$$

but this is equal to neither of the two equations... so now i have three equations.. which one is correct??

Last edited: Feb 24, 2014
4. Feb 24, 2014

rude man

I won't deal with the E field. Much too laborious.

Anyway, your equation here is where I'd put my money. Integrated from 0 to L.

Last edited: Feb 24, 2014