Force between plates of a capacitor

Consider two line charges at a distance x from bottom oon plate 1, and a distance y from bottom on plate 2. From Find the force of interaction-2 , we know that horizontal component of force between these wires is: $dF_{12} =\displaystyle \dfrac{\lambda^2}{2 \pi \epsilon_{0} \sqrt{l^2+(y-x)^2}}\bigg(\sqrt{(y-x)^2+l^2+l^2} - \sqrt{(y-x)^2+l^2}\bigg) \dfrac{l}{\sqrt{l^2+(y-x)^2}}$

Integrating, we get:

$F = \displaystyle \int_{0}^{l} \int_{0}^{l} \dfrac{\sigma^2 l {\mathrm dx} {\mathrm dy}}{2 \pi \epsilon_{0} (l^2+(y-x)^2)}\bigg(\sqrt{(y-x)^2+2l^2} - \sqrt{(y-x)^2+l^2}\bigg)$

= $\displaystyle \dfrac{\sigma^2 l} {2 \pi \epsilon_{0} } \displaystyle \int_{0}^{l} \int_{0}^{l} \dfrac{ \sqrt{(y-x)^2+2l^2} - \sqrt{(y-x)^2+l^2} }{ (l^2+(y-x)^2)}{\mathrm dx} {\mathrm dy}$

It can be solved by first integrating wrt y, and then x,all integrals present are trivial, but solution is quite lengthy.

This is not a complete solution as I did not compute the final double integral by hand (I used Wolfram Alpha). However, what I do show is how we can reduce the quadruple integral into the double integral.

Let the first plate be described by the region

$0\leq x\leq l,0\leq y\leq l, z=0$

and the second plate be described by the region

$0\leq x'\leq l,0\leq y'\leq l, z'=l$

It can be easily shown by Coulomb's law that the force between the plates is given by

$\mathbb{F}=\dfrac{\sigma^{2}l}{4\pi\epsilon_{0}}\displaystyle\int_{0}^{l}\!\!\!\displaystyle\int_{0}^{l}\!\!\!\displaystyle\int_{0}^{l}\!\!\!\displaystyle\int_{0}^{l} \frac{\text{d}x\text{ }\text{d}x'\text{d}y\text{ }\text{d}y'}{\left( (x-x')^{2}+(y-y')^{2}+l^{2}\right)^{3/2}}\hat{z}$

Making the substitution

$x=Xl,x'=X'l,y=Yl,y'=Y'l$

$\mathbb{F}=\dfrac{\sigma^{2}l^{2}}{4\pi\epsilon_{0}}\displaystyle\int_{0}^{1}\!\!\!\displaystyle\int_{0}^{1}\!\!\!\displaystyle\int_{0}^{1}\!\!\!\displaystyle\int_{0}^{1} \frac{\text{d}X\text{ }\text{d}X'\text{d}Y\text{ }\text{d}Y'}{\left( (X-X')^{2}+(Y-Y')^{2}+1\right)^{3/2}}\hat{z}$

By symmetry we may also integrate over the region

$0\leq X\leq 1, X'\leq X$

$0\leq Y\leq 1, Y'\leq Y$

and multiply by $4$ .

Now to compute the integral make the substitution

$X=\dfrac{u+v}{2},X'=\dfrac{u-v}{2}$

$Y=\dfrac{a+b}{2},Y'=\dfrac{a-b}{2}$

$\therefore \mathbb{F}=\dfrac{\sigma^{2}l^{2}}{4\pi\epsilon_{0}}\displaystyle\int_{0}^{1}\!\!\!\displaystyle\int_{b}^{2-b}\!\!\!\displaystyle\int_{0}^{1}\!\!\!\displaystyle\int_{v}^{2-v} \frac{\text{d}u\text{ }\text{d}v\text{ }\text{d}a\text{ }\text{d}b}{\left( v^{2}+b^{2}+1\right)^{3/2}}\hat{z}$

Note that the factor of $4$ that was multiplied earlier cancels with the absolute value of the determinant of the Jacobian (which is $1/4$ ).

$\therefore \mathbb{F}=\dfrac{\sigma^{2}l^{2}}{\pi\epsilon_{0}}\displaystyle\int_{0}^{1}\!\!\!\displaystyle\int_{0}^{1} \frac{(1-v)(1-b)\text{d}v\text{ }\text{d}b}{\left( v^{2}+b^{2}+1\right)^{3/2}}\hat{z}$

I computed this integral on Wolfram Alpha. After plugging in the values we get

$\mathbb{F}=1.000004255\hat{z}$

$\therefore \boxed{\mathbb{F}\approx 1\hat{z} N}$

Closed form

$F= \dfrac{{\sigma}^{2}{l}^{2}} {\pi{\epsilon}_{0}} \left( 2ln \left( \dfrac{1+\sqrt{3}}{1+\sqrt{2}}\right) + \dfrac{\pi}{6} + \sqrt{8}-(\sqrt{3}+1+ln(2)) \right)$

its a good question it take 7 min to do

Great question and solutions but it seems that the approximation holds well enough and gives answer as 1.0189 . 😀😁😂😝😜