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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 $\sigma$ and $-\sigma$.

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

 Problem Figure description

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 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 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 $2$

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 $x=b$.

Any other methods to find the field, I tried with polar coordinates but I don't know how to do it.

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

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You can get the field for the tape by considering an infinite sum of parallel line charges.

$$E(a)=\int_{r=0}^{r=b}E_{line}(a+r)dr$$

$E_{line}$ is pretty standard result so I'll assume it.

$$ E(a)=\int_{r=0}^{r=b}\frac{\lambda}{2\pi\epsilon_0(a+r)}dr $$ The line charge density for each line will be the surface charge density $\sigma$ divided by the width $b$. Integrating: $$ E(a)=\left[\frac{\sigma\ln(a+r)}{2\pi\epsilon_0b}\right]_{r=0}^{r=b} $$

$$ E(a)=\frac{\sigma}{2\pi\epsilon_0b}\left[\ln(a+b)-\ln(a)\right] $$

Which, with a little massaging, agrees with your expression for the field produced by the right tape a distance $a$ in the same plane from the edge of the tape

$$ E=k\sigma\ln\left( \frac{(a+b)^2}{a^2} \right) $$

Force per unit height: The left tape sits in the field of the right tape. An infinitesimal of charge in the left tape $\mathrm{d}q$ feels an infinitesimal force $\mathrm{d}F=E\mathrm{d}q$. Integrating up $F=\int E\mathrm{d}q=\int E\rho\mathrm{d}A$. Using $\rho=\sigma$ and appropriate boundaries:

$$F=\int_{y=0}^{y=1}\int_{a=0}^{a=b} k\sigma^2\ln\left( \frac{(a+b)^2}{a^2} \right)\mathrm{d}a\mathrm{d}y$$

$$F=2k\sigma^2\int_{a=0}^{a=b} [\ln (a+b) -\ln (a)] \mathrm{d}a$$ Integrating by Parts $$F=2k\sigma^2\left[ a\ln(a+b)-a\ln(a) \right]_{a=0}^{a=b} -2k\sigma^2\int_{a=0}^{a=b} \left[a\frac{1}{a+b} -a\frac{1}{a}\right] \mathrm{d}a$$ Letting $a+b=c$ $$F=2k\sigma^2b\ln(2) +2k\sigma^2b -2k\sigma^2\int_{c=b}^{c=2b} \frac{c-b}{c} \mathrm{d}c$$

$$F=2k\sigma^2b\ln(2) +2k\sigma^2b +2k\sigma^2\left[ b\ln(c) -c\right]_{c=b}^{c=2b}$$

$$F=2k\sigma^2b\ln(2) +2k\sigma^2b +2k\sigma^2\left[b\ln(2)-b \right]$$

$$ F=4k\sigma^2b\ln(2) $$

which is the given answer.

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  • $\begingroup$ In accordance with our homework policy, I'm temporarily deleting this. $\endgroup$ – Qmechanic Sep 20 '14 at 13:54

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