# Archived Find the field of two infinite strips of width b

1. Sep 20, 2014

### KeithPhysics

> Q:The two charged strips in the following picture have width $b$, infinite height,and negligible thickness(in the direction perpendicular to the page).Their densities per unit area are $\pm \sigma$ .​

>a)Find the field magnitude produced due to one of the strips from a distance $a$ away from it(in the plane of the page).​

>b)Show that the force(per unit height) between the two strips is equal to $\frac{\sigma^2 b \ln(2)}{\pi \epsilon_0}$. Note that this result is finite, even tough the field in the boundary of the strip diverges.​

>*Part a) solved in the following lines:

>**Struggling with part b)

Attempted Solution:

>Part a) solved in the following lines:

As the figure shows I calculate the Field from a distance $a$ from the first tape:

\begin{aligned}E_x&=\int\int \frac{k dq}{r^2}\cos{\theta} \\ &=k\int\int \frac{\sigma da}{r^2}\cos{\theta} \end{aligned}

From the geometry of the figure we get:

\begin{aligned}E_x&=k\sigma\int_{-\infty}^{\infty}\int_0^b \frac{ dx dy}{(a+x)^2+y^2}\cdot\frac{a+x}{\sqrt{(a+x)^2+y^2}} \\ &=k\sigma\int_{-\infty}^{\infty}\Big(\int_0^b \frac{ (a+x)dx }{\Big((a+x)^2+y^2\Big)^{3/2}}\Big)dy\end{aligned}

If $u= (a+x)^2+y^2=(a^2+2ax+x^2+y^2)$, thus $du=(2a+2x)dx \Rightarrow du/2=(a+x)dx$

\begin{aligned}E_x&=k\sigma\int_{-\infty}^{\infty}\Big( \int_{c'}^{d'} \frac{du}{2u^{3/2}} \Big)dy=\frac{k\sigma}{2}\int_{-\infty}^{\infty}\Big( \Big[ \frac{-2}{u^{1/2}}\Big]_{c'}^{d'} \Big)dy \\ &=-k\sigma \int_{-\infty}^{\infty}\Big( \Big[(a+x)^2+y^2)^{-1/2} \Big]_{0}^{b} \Big)dy \\ &=-k\sigma \int_{-\infty}^{\infty}\Big( (a+b)^2+y^2)^{-1/2} -(a^2+y^2)^{-1/2} \Big)dy \\ &=-k\sigma \int_{-\infty}^{\infty}\Big( \frac{1}{\sqrt{(a+b)^2+y^2}} - \frac{1}{\sqrt{a^2+y^2}} \Big)dy \end{aligned}

We can find an anti-derivative for the previous expression:
\begin{aligned}E_x&=-k\sigma \Big[ \ln(\sqrt{(a+b)^2+y^2}+y)-\ln(\sqrt{a^2+y^2}+y)\Big]_{-\infty}^{\infty}\\ &=-k\sigma \lim_{d \rightarrow \infty} \Big[ \ln\Big(\frac{\sqrt{(a+b)^2+y^2}+y}{\sqrt{a^2+y^2}+y}\Big) \Big]_{-d}^{d} \\ &=-k\sigma \lim_{d \rightarrow \infty} \Big[ \ln\Big(\frac{\sqrt{(a+b)^2+d^2}+d}{\sqrt{a^2+d^2}+d}\Big) - \ln\Big(\frac{\sqrt{(a+b)^2+d^2}-d}{\sqrt{a^2+d^2}-d}\Big)\Big] \end{aligned}

The first limit can be easily seen as tending to $0$ and the second one according to [Wolfram] https://www.wolframalpha.com/input/?i=lim+d+->infinity++log((d-sqrt((c)^2+d^2))/(d-sqrt((f)^2+d^2))) tends to $\ln(\frac{(a+b)^2}{a^2})$ something that is rather convenient.

So the intensity of the field of the $\sigma$ tape is equal to :

$$E_{\sigma}=-k\sigma \ln \Big(\frac{(a+b)^2}{a^2}\Big)$$

This result seems right because the field diverges in $0$ and in $-b$ .

Using a similar approach we can calculate the other plate field changing the limits of integration of $x$ from $b$ to $2b$

\begin{aligned}E_x&=k\sigma\int_{-\infty}^{\infty}\Big(\int_b^{2b} \frac{ (a+x)dx }{\Big((a+x)^2+y^2\Big)^{3/2}}\Big)dy \\ &=-k\sigma \int_{-\infty}^{\infty}\Big( \Big[(a+x)^2+y^2)^{-1/2} \Big]_b^{2b} \Big)dy \\ &=-k\sigma \int_{-\infty}^{\infty}\Big( (a+2b)^2+y^2)^{-1/2} -((a+b)^2+y^2)^{-1/2} \Big)dy \\ &=-k\sigma \int_{-\infty}^{\infty}\Big( \frac{1}{\sqrt{(a+2b)^2+y^2}} - \frac{1}{\sqrt{(a+b)^2+y^2}} \Big)dy \end{aligned}

We can find an anti-derivative for the previous expression:
\begin{aligned}E_x&=-k\sigma \Big[ \ln(\sqrt{(a+2b)^2+y^2}+y)-\ln(\sqrt{(a+b)^2+y^2}+y)\Big]_{-\infty}^{\infty}\\ &=-k\sigma \lim_{d \rightarrow \infty} \Big[ \ln\Big(\frac{\sqrt{(a+2b)^2+y^2}+y}{\sqrt{(a+b)^2+y^2}+y}\Big) \Big]_{-d}^{d} \\ &=-k\sigma \lim_{d \rightarrow \infty} \Big[ \ln\Big(\frac{\sqrt{(a+2b)^2+d^2}+d}{\sqrt{(a+b)^2+d^2}+d}\Big) - \ln\Big(\frac{\sqrt{(a+2b)^2+d^2}-d}{\sqrt{(a+b)^2+d^2}-d}\Big)\Big] \end{aligned}

Again the first limit can be easily seen as tending to $0$ and the second one according to [Wolfram] https://www.wolframalpha.com/input/?i=lim+d+->infinity++log((d-sqrt((c)^2+d^2))/(d-sqrt((f)^2+d^2))) tends to $\ln(\frac{(a+2b)^2}{(a+b)^2})$ something that is again rather convenient.

So the intensity of the field of the $-\sigma$ tape is equal to :

$$E_{-\sigma}=k\sigma \ln \Big(\frac{(a+2b)^2}{(a+b)^2}\Big)$$

This result seems right because the field diverges in $-b$ and in $-2b$ .

Now if I sum the two fields:

$$E_T=E_{-\sigma}+E_{-\sigma}=-k\sigma \ln \Big(\frac{(a+b)^2}{b^2}\Big) + k\sigma \ln \Big(\frac{(a+2b)^2}{(a+b)^2}\Big)$$
Reducing the last expression and using the properties of logarithms:
$$E_T=-2k\sigma \ln \Big(\frac{(a+b)}{b}\Big) + 2k\sigma \ln \Big(\frac{(a+2b)}{(a+b)}\Big)=-2k\sigma \ln \Big(\frac{a(a+2b)}{(a+b)^2}\Big)$$

It diverges when $x=b$.

I tried with polar coordinates and I arrived to the same solution but much easier.

>How can solve part b), I don't know how to do it.​

2. Feb 4, 2016

### Staff: Mentor

There is no need to sum fields. The force on the left strip comes from the right force only (forces from the left strip on the left strip cancel due to symmetry).

That just leaves an integral over the electric field strength multiplied by the charge density, within the borders of the strip. That integral looks easy to evaluate.