Using the gaussian law I want to determine the field strength of a charged stick on another load.
The base formula for that is $$\epsilon_0 \oint {\vec{E} \cdot d \vec{A}} = q$$ A second formula for q is:
$$ q = \lambda \cdot h $$ $$\vec{E} = \frac{1}{4\cdot \pi \cdot \epsilon_0} \cdot \frac{q_1}{r^2} \cdot \hat{r}$$ In this case we use a cylinder to solve our problem and we only need the lateral surface.
$$A = 2\cdot \pi \cdot d \cdot h$$

My questions

  1. I have a strong problem with a surface written as vector.
    Assuming we have a situation like in the first picture of this question where the lateral surface of our cylinder touches the load.
    How would I properly write the surface as vector (how can I determine the direction vector)?
    Usually my approach would be $\vec{r} = (x_2 - x_1)\hat{x} + (y_2 - y_1)\hat{y}$. But as we have a solid here I have problems finding the start and end coordinates.

  2. Until now I only solved integrals in the form of $\int f(x) dx$ but now I have an equation like the one above. What would be my "x" here? How would I solve this integral?

As you can imagine this is a task from my electronics class. It's not a homework but it was an example.
Unfortunately we did explicitly not pay attention to the integral and we did not pay attention to the surface written as vector.
What we did is the following:
$$\epsilon_0 \oint {\vec{E} \cdot d \vec{A}} = q$$ $$\epsilon_0 \cdot |\vec{E}| \cdot 2 \cdot \pi \cdot h \cdot d = q$$ Because $\lambda$ is given in this particular task we can do $$\epsilon_0 \cdot |\vec{E}| \cdot 2 \cdot \pi \cdot h \cdot d = \lambda \cdot h$$ $$|\vec{E}| = \frac{\lambda}{2\cdot \pi \cdot d \cdot \epsilon_0}$$


1 Answer 1


Due to symmetry considerations, $\vec E$ at a given point is always pointing along the line from that point that is perpendicular to the wire. We can use cylindrical coordinates. With $d$ the distance from the wire, we can write $\vec E=E(d)\hat r$. For a surface, the vector $d\vec A$ is always perpendicular to that surface. So in cylindrical coordinates we can write the surface element that is part of the cylinder with radius $r$ as $d\vec A=r d\theta dh \hat r$. Notice that $d\vec A$ and $\vec E$ are parallel (or antiparallel), so $$\vec E\cdot d\vec A=EdA=E d\ d\theta dh$$ Integrating $\theta$ from $0$ to $2\pi$ and $h$ from $0$ to $h$, you get your formula. In conclusion, you get the answer due to the fact that $\vec E$ is always perpendicular to the surface of the cylinder of radius $d$, and is only dependent on the magnitude of $d$.

  • $\begingroup$ Thanks again! I have still one question: How can I find the antiderivative of this? As I mentioned, unfortunately we skipped this part. I know that this is probably absolutely simple but it confuses me anyway. Secondly another probably quite silly question: How would the unit vectors look like? $\endgroup$
    – TimSch
    Nov 17, 2018 at 23:37
  • 1
    $\begingroup$ Integral over $d\theta$ is $2\pi$, integral over $dh$ is the length of the wire $L$ (will go to infinity, but will cancel with the charge: $q=\lambda L$) $\endgroup$
    – Andrei
    Nov 17, 2018 at 23:59

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