How Does Linear Charge Density Affect Electric Field and Potential at a Point?

[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..

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

--------------------------

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..

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

 

[rude man]

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

Your last integral does not make sense ...

 

[iScience]

i added the dx..

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??

 

[rude man]

i added the dx..

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??

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.