I know the formula of the electric field, however:

suppose that we put two infinitely long and thin, straight wires symetrically into the coordinate system, so that y axis is between them.

Now consider that we have wires charged with charge $q_1=q$ and $q_2= -q$.

enter image description here

If we sum the contributions to electric field from both wires with vector summation (note that the field has only radial component) in point T (T lies anywhere on the X=0 plane), we find out that the distance from both wires to the point T where we want to calculate the electric field, is the same.

But due to the fact that we have a positive and negative charge, i get:

(this is the part of the equation)

$(\frac{1}{r} - \frac{1}{r}) = 0$; where $r_1=r_2=r$

enter image description here

$$E_1 = \frac{q}{2\pi\varepsilon_0\rho_1}$$

$$E_2 = \frac{-q}{2\pi\varepsilon_0\rho_2}$$

which means that electric field on plane x=0 is zero, but I know that it isn't, because it is perpendicular to the equipotential V=0.

So where do I go wrong in calculating the field in point T?

In the end I would like to show that the electric field is indeed perpendicular to the equipotential V=0 which is on x=0, but i can't even start doing that when I get that electric field is 0 there....

  • $\begingroup$ Although you're using commonly known equations, could you still post the full form here and maybe add a sketch of the geometry? It just makes it easiser to help you. $\endgroup$
    – engineer
    Commented May 20, 2015 at 10:04
  • $\begingroup$ It seems you're assuming that the wires have electric fields $E=\pm\frac{\lambda}{2\pi r\epsilon_0}\hat{r}$ with $\lambda$ the charge density. However, then the unit vector $\hat{r}$ is with respect to cylindrical coordinates with the wires placed at $z=0$. Since the wires are at different coordinates $z$, the two unit vectors $\hat{r}$ defined with respect to the two wires point in different directions in the point $T$. $\endgroup$ Commented May 20, 2015 at 10:15
  • $\begingroup$ @engineer and ScroogeMcDuck: Here, I uploaded sketch of the geometry :P Hopefully you can see what i mean :) i62.tinypic.com/fxy0za.png also here is the equation i57.tinypic.com/2ue4d3a.png $\endgroup$
    – antestor
    Commented May 20, 2015 at 11:16

1 Answer 1


I think your problem is that you're adding the electric fields like scalars, rather than breaking then into their vector components. You have:

$ E_1 = \frac{q}{2\pi\varepsilon_0\rho_1} $ and $ E_2 = \frac{-q}{2\pi\varepsilon_0\rho_2} $

What you should have is

$ E_1 = \frac{q}{2\pi\varepsilon_0}(\frac{1}{\rho_{1x}}\hat{x} + \frac{1}{\rho_{1y}}\hat{y})$ and $ E_2 = \frac{-q}{2\pi\varepsilon_0}(\frac{1}{\rho_{2x}}\hat{x} + \frac{1}{\rho_{2y}}\hat{y})$

Where we have split the radial distances $\rho_1$ and $\rho_2$ into their $x$- and $y$-components.

If we note that $\rho_{1x} = -\rho_{2x} = \rho_x$ and $\rho_{1y} = \rho_{2y} = \rho_y$ this becomes:

$ E_1 = \frac{q}{2\pi\varepsilon_0}(\frac{1}{\rho_x}\hat{x} + \frac{1}{\rho_y}\hat{y})$ and $ E_2 = \frac{-q}{2\pi\varepsilon_0}(\frac{1}{-\rho_x}\hat{x} + \frac{1}{\rho_y}\hat{y})$

Then adding the two vectors together the $y$-components cancel (due to the opposite charge but equal position vectors) and you're left with the x-components only:

$ E_1 + E_2 = \frac{q}{2\pi\varepsilon_0}(\frac{2}{\rho_x}\hat{x}) = \frac{q}{\pi\varepsilon_0\rho_x}\hat{x}$

i.e. the electric field is only in the $x$-direction, as required.

  • $\begingroup$ Yes, that does seem to solve my question, thank you very much :) can't believe i overlooked that.. $\endgroup$
    – antestor
    Commented May 20, 2015 at 13:41

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