Finding voltage from the E field

1. Feb 19, 2014

iScience

consider..

If i'm trying to find the voltage at point P due to a linear charge density of length L (the white line that lies on the x axis):

then how come it is incorrect to do the following?

$$V=\int{\vec{E_{x}}dx}+\int{\vec{E_{z}}dz}$$?

(the r vector is just to find dE at each point on the charge density of length L)

Last edited: Feb 19, 2014
2. Feb 19, 2014

BvU

Hello I,

What exactly does your equation mean ? Ex ? Ez ? what is dx, dz and what are the integration bounds ?

Also: is your endering of the exercise complete ? r ?

3. Feb 19, 2014

iScience

$$\vec{E_{x}}$$ and $$\vec{E_{z}}$$ are the electric field vectors in the specified directions, x, and z respectively.

I'm not concerned with specific bounds at this time right now i am only concerned with whether or not i have the right expression for potential at some point in space i decided to call "P".

i don't know what you mean

"(the r vector is just to find dE at each point on the charge density of length L)"

4. Feb 20, 2014

BvU

You want to find the potential at point P due to the line charge on the x-axis.

If you already, at all points in space, know the electric field, cause by the line charge, then you can integrate, e.g. Ez dz from z =∞ to z = z(P) to get V.

Or Ex dx from x =-∞, z = z(P) to x = 0, z = z(P)

These two under the assumption that V = 0 far away from the wire (which x = ∞, z = z(P) does not satisfy).

They look like your integrals, only one has dx = 0 and the other dz = 0 (makes life easier).

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My question about the Rendering (sorry) of the original problem was inspired by the fact that I am missing an indication of what z is, what r is, whether the charge density on the line is a constant, ...

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Since you don't work out your integrals, I am inclined to believe you do not have an expression for $\vec E(x,z)$ for all $x,z$. You could embark on finding one, but the exercise only asks for the potential.

I take it you do have an expression for V as a function of q and r at hand. So if you chop your line charge into little pieces with length dx, each little piece having a charge $dq = \lambda dx$ and contributing a little $dV = ... dq = ... dx$, I am sure you can easily write down an integral that finds you V diectly!