Electric field due to a uniformly charged FINITE rectangular plate

Question

I was teaching kids about how to find electric field using the superposition principle for continuous charge distributions. I thought maybe I should derive the formula for electric field due to a finite rectangular sheet of charge of charge on the surface $S$, where $$ S = \left\{(x,y,z)\in \mathbb{R}^3 \mid -a/2< x < +a/2; -b/2< y < +b/2 ; z = 0 \right\} .$$ However, I got stuck at the following integration. $$ E(0,0,r) = \frac{\sigma r}{4\pi\epsilon_o} \int_{x=-a/2}^{x=+a/2}\int_{y=-b/2}^{y=+b/2} \frac{dx dy}{(x^2+y^2+r^2)^{3/2}}, $$ where $\sigma$ is the surface charge density.

Note: This integration can be done if $a$ or $b$ or both are very large i.e. $\infty$ in which case we get usual result of $E=\frac{\sigma}{2\epsilon_o}$

So my question is, Can this integral be calculated? If not then what method would I use to find the electric field in this case. Also It would be great if anyone can comment on how to find the electric field by directly solving the poisson equation.

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Consequently if we take case of finite disk the following is the resulting integration.

$$ E = \frac{\sigma r}{2\epsilon_o} \int_{\xi=0}^{\xi=R} \frac{\xi d\xi}{(\xi^2+r^2)^{3/2}} $$

which can be solved as

$$ E = \frac{\sigma}{2\epsilon_o} \left(1- \frac{r}{\sqrt{r^2+R^2}}\right) $$

Now by taking the limit $R \rightarrow \infty$ we can show that $E \rightarrow \frac{\sigma}{2\epsilon_o}$.

Answer 1

The integrals are difficult but not impossible, unless I've made a mistake with WolframAlpha. The result is:

$$E = \frac{\sigma}{\pi \epsilon_0} \arctan\left( \frac{ab}{4r\sqrt{(a/2)^2+(b/2)^2+r^2}} \right)$$

When $a,b \to \infty$ the whole arctangent goes to $\pi/2$ and we recover $E=\frac{\sigma}{2\epsilon_0}$, which is definitely encouraging.

And I don't know what you mean by "directly solving Poisson's equation". As far as I know, the usual way to do that is to use Green's functions, i.e., this integral.

Answer 2

initially put $x^2 + r^2 = p^2$

then $y = p(\tan(A))$

solve it, it will be in terms in terms of $x$. and substitute $\frac{r(\tan(B))}{b} = \frac{x}{\sqrt{4x^2 + 4r^2 + b^2}}$ solve it you will get answer easily.

Answer 3

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In Figure-01 we have a finite rectangular plate with uniform surface charge density $\:\sigma\:$ arranged symmetrically with respect to the $\:\mr x-$axis and $\:\mr y-$axis separately. We want to find the electric field intensity vector $\:\mb E\:$ at the field point $\:\texttt P\e\plr{0,0,h}\:$ on the $\:\mr z-$axis.

Consider an infinitesimal rectangular surface element $\:\mr dx\mr dy\:$ around the point $\:\texttt A\e\plr{x,y,0}\:$ as shown in Figure-02. For the infinitesimal electric field intensity vector $\:\mr d\mb E\:$ at the field point $\:\texttt P\e\plr{0,0,h}\:$ we have \begin{equation} \mr d\mb E \e \plr{\mr{dE}_x,\mr{dE}_y,\mr{dE}_z}\e\dfrac{\mr d q}{4\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \dfrac{\mb p}{\:\,\Vlr{\mb p}^3}\e \dfrac{\sigma\mr dx\mr dy}{4\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \dfrac{\mb p}{\:\,\Vlr{\mb p}^3} \tl{01} \end{equation} where $\:\mb p \e\plr{x,y,h}\:$ the position vector of point $\:\texttt P\:$ with respect to the infinitesimal surface element $\:\mr dx\mr dy$.

Integration of the components $\:\mr{dE}_x\:$ and $\:\mr{dE}_y\:$ on the plate surface would give null results due to the mirror symmetry with respect to the $\:\mr{yz}\:$ and $\:\mr{xz}\:$ planes respectively. For the $\:\mr{dE}_z\:$ component

\begin{equation} \mr{dE}_z \e \dfrac{\sigma\mr dx\mr dy}{4\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \dfrac{\cos\theta}{\:\,\Vlr{\mb p}^2}\e \dfrac{\sigma\mr dx\mr dy}{4\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \dfrac{h}{\:\,\Vlr{\mb p}^3} \e \dfrac{\sigma h}{4\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \dfrac{\mr dx\mr dy}{\plr{x^2\p y^2\p h^2}^{3/2}} \tl{02} \end{equation} that is

\begin{equation} \mr {dE}_z \e \dfrac{\sigma h}{4\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \dfrac{\mr dx\mr dy}{\plr{x^2\p y^2\p h^2}^{3/2}} \tl{03} \end{equation} so

\begin{equation} \mr E_z \e \dfrac{\sigma h}{4\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \int\limits_{y\e \m b}^{y\e \p b}\blr{\:\:\int\limits_{x\e \m a}^{x\e \p a}\!\!\dfrac{\mr dx}{\plr{x^2\p y^2\p h^2}^{3/2}}}\mr dy \tl{04} \end{equation} From the indefinite integral^(A)

\begin{equation} \int\dfrac{\mr dx}{\plr{x^2\p A^2}^{3/2}\Vp{\dfrac12}}\e \dfrac{x}{A^2\sqrt{x^2\p A^2\Vp{\tfrac12}}}\p \texttt{constant} \tl{05} \end{equation} we have \begin{equation} \begin{split} \int\limits_{x\e \m a}^{x\e \p a}\!\!\dfrac{\mr dx}{\plr{x^2\p y^2\p h^2}^{3/2}} & \e \blr{\dfrac{x}{\plr{y^2\p h^2}\sqrt{x^2\p y^2\p h^2\Vp{\tfrac12}}}}_{x \e \m a}^{x \e \p a} \\ & \e \dfrac{2a}{\plr{y^2\p h^2}\sqrt{y^2\p a^2\p h^2\Vp{\tfrac12}}} \end{split} \tl{06} \end{equation} so equation \eqref{04} yields

\begin{equation} \mr E_z \e \dfrac{\sigma a h}{2\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \int\limits_{y\e \m b}^{y\e \p b}\dfrac{\mr dy}{\plr{y^2\p h^2}\sqrt{y^2\p a^2\p h^2\Vp{\tfrac12}}} \tl{07} \end{equation} From the indefinite integral^(B)

\begin{equation} \int\dfrac{\mr dy}{\plr{y^2\p A^2}\sqrt{y^2\p A^2\p B^2\Vp{\tfrac12}}}\e \dfrac{\arctan{\plr{\dfrac{By}{A\sqrt{y^2\p A^2\p B^2\Vp{\tfrac12}}}}}}{AB\Vp{\Vlr{\mb p}^3}}\p \texttt{constant} \tl{08} \end{equation} we have

\begin{equation} \begin{split} \int\limits_{y\e \m b}^{y\e \p b}\dfrac{\mr dy}{\plr{y^2\p h^2}\sqrt{y^2\p a^2\p h^2\Vp{\tfrac12}}} & \e\dfrac{\blr{\arctan{\plr{\dfrac{ay}{h\sqrt{y^2\p h^2\p a^2\Vp{\tfrac12}}}}}}_{y\e\m b}^{y\e\p b}}{ah\Vp{\Vlr{\mb p}^3}}\\ & \e \dfrac{2\arctan{\plr{\dfrac{ab}{h\sqrt{a^2\p b^2\p h^2\Vp{\tfrac12}}}}}}{ah\Vp{\Vlr{\mb p}^3}}\\ \end{split} \tl{09} \end{equation} so equation \eqref{07} yields

\begin{equation} \boxed{\:\:\mr E_z \e \dfrac{\sigma}{\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \arctan{\plr{\dfrac{ab}{h\sqrt{a^2\p b^2\p h^2\Vp{\tfrac12}}}}}\e\dfrac{\sigma}{\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \Theta\e \dfrac{\sigma}{4\pi\epsilon_0\Vp{\Vlr{\mb p}^3}} \Omega\:\:\Vp{\tfrac{\dfrac{\dfrac{e}{f}}{b}}{\dfrac{c}{\dfrac{e}{f}}}}} \tl{10} \end{equation} where $\:\Theta\:$ the tetrahedral solid angle at point $\:\texttt{P}$^(C) of the tetrahedron with vertex $\:\texttt{P}\:$ and base anyone of the four rectangles $\: a\times b$, see Figure-03, and $\:\Omega\e 4\Theta\:$ the tetrahedral solid angle at point $\:\texttt{P}\:$ of the tetrahedron with vertex $\:\texttt{P}\:$ and base the rectangle $\:\texttt Q_1\texttt Q_2\texttt Q_3\texttt Q_4$, see Figure-04.

Note that
\begin{equation} \lim_{a,b\,\bl\rightarrow\bl\infty}\!\!\!\Theta\e \dfrac{\pi}{2}\,\quad \bl\implies \quad \lim_{a,b\,\bl\rightarrow\bl\infty}\!\!\!\mr E_z\e \dfrac{\sigma}{2\epsilon_0\Vp{\Vlr{\mb p}^3}} \tl{11} \end{equation} the well-known result for the infinite plane.

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^{(A) Proof of equation \eqref{05} \begin{equation} \int\dfrac{\mr dx}{\plr{x^2\p A^2}^{3/2}\Vp{\dfrac12}}\e \dfrac{x}{A^2\sqrt{x^2\p A^2\Vp{\tfrac12}}}\p \texttt{constant} \tl{A-01} \end{equation} Make variable change \begin{equation} x\e A\tan u \tl{A-02} \end{equation} so \begin{equation} \mr dx \e \dfrac{A}{\cos^2 u\Vp{\tfrac12}}\mr du\,,\quad \dfrac{1}{\plr{x^2\p A^2}^{3/2}\Vp{\tfrac12}}\e \dfrac{\cos^3 u}{A^3\Vp{\tfrac12}} \tl{A-03} \end{equation} and \begin{equation} \begin{split} \int\dfrac{\mr dx}{\plr{x^2\p A^2}^{3/2}\Vp{\tfrac12}}&\e \int\dfrac{\cos u\mr du}{A^2\Vp{\tfrac12}}\e \int\dfrac{\mr d\plr{\sin u}}{A^2\Vp{\tfrac12}}\\ & \e \dfrac{\sin u}{A^2\Vp{\tfrac12}}\p\texttt{constant}\e\dfrac{x}{A^2\sqrt{x^2\p A^2\Vp{\tfrac12}}}\p \texttt{constant} \end{split} \tl{A-04} \end{equation} since from \eqref{A-02} \begin{equation} \begin{split} \sin u \e \dfrac{\tan u}{\sqrt{1\p \tan^2 u\Vp{\tfrac12}}}\e\dfrac{\plr{x/A}}{\sqrt{1\p \plr{x/A}^2\Vp{\tfrac12}}}\e\dfrac{x}{\sqrt{x^2\p A^2\Vp{\tfrac12}}} \end{split} \tl{A-05} \end{equation}}

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^{(B) Proof of equation \eqref{08} \begin{equation} \int\dfrac{\mr dy}{\plr{y^2\p A^2}\sqrt{y^2\p A^2\p B^2\Vp{\tfrac12}}}\e \dfrac{\arctan{\plr{\dfrac{By}{A\sqrt{y^2\p A^2\p B^2\Vp{\tfrac12}}}}}}{AB\Vp{\Vlr{\mb p}^3}}\p \texttt{constant} \tl{B-01} \end{equation} Make variable change \begin{equation} y\e \sqrt{A^2 \p B^2\Vp{\tfrac12}}\tan v \tl{B-02} \end{equation} so \begin{equation} \begin{split} \mr dy & \e \dfrac{\sqrt{A^2 \p B^2\Vp{\tfrac12}}}{\cos^2 v\Vp{\tfrac12}}\mr dv\,,\quad \sqrt{y^2\p A^2\p B^2\Vp{\tfrac12}}\e \dfrac{\sqrt{A^2 \p B^2\Vp{\tfrac12}}\Vp{\tfrac12}}{\cos v}\\ y^2\p A^2 & \e \dfrac{B^2\sin^2 v\p A^2}{\cos^2 v} \end{split} \tl{B-03} \end{equation} and \begin{equation} \begin{split} &\dfrac{\mr dy}{\plr{y^2\p A^2}\sqrt{y^2\p A^2\p B^2\Vp{\tfrac12}}} \e \dfrac{\plr{\dfrac{\sqrt{A^2 \p B^2\Vp{\tfrac12}}}{\cos^2 v\Vp{\tfrac12}}}\mr dv}{\plr{\dfrac{B^2\sin^2 v\p A^2}{\cos^2 v}\Vp{\dfrac{\sqrt{A^2 \p B^2\Vp{\tfrac12}}\Vp{\tfrac12}}{\cos v}}}\plr{\dfrac{\sqrt{A^2 \p B^2\Vp{\tfrac12}}\Vp{\tfrac12}}{\cos v}}}\\ &\e \dfrac{\cos v \mr dv}{B^2\sin^2 v\p A^2}\e \dfrac{1}{AB}\dfrac{\mr d\plr{\dfrac{B}{A}\sin v}}{1\p\plr{\dfrac{B}{A}\sin v}^2}\e \dfrac{1}{AB\Vp{{A^2}^2}}\dfrac{\mr d \mr w}{\plr{1\p\mr w^2}\Vp{{A^2}^2}}\\ &\e \dfrac{1}{AB}\mr d\plr{\arctan \mr w\Vp{A^2}} \end{split} \tl{B-04} \end{equation} that is \begin{align} &\dfrac{\mr dy}{\plr{y^2\p A^2}\sqrt{y^2\p A^2\p B^2\Vp{\tfrac12}}}\e \dfrac{1}{AB}\mr d\plr{\arctan \mr w\Vp{A^2}} \tl{B-05a}\\ &\texttt{where} \nonumber\\ &\mr w \e \dfrac{B}{A}\sin v\e \dfrac{B}{A}\dfrac{\tan v}{\sqrt{1 \p \tan^2 v\Vp{\tfrac12}}}\e \dfrac{B y}{A\sqrt{y^2\p A^2\p B^2\Vp{\tfrac12}}} \tl{B-05b} \end{align} From equations \eqref{B-05a} and \eqref{B-05b} we have \begin{equation} \begin{split} &\int\dfrac{\mr dy}{\plr{y^2\p A^2}\sqrt{y^2\p A^2\p B^2\Vp{\tfrac12}}}\e \dfrac{1}{AB}\int\mr d\plr{\arctan \mr w\Vp{A^2}}\e \dfrac{\arctan \mr w}{AB}\p\texttt{constant}\\ &\e\dfrac{\arctan{\plr{\dfrac{By}{A\sqrt{y^2\p A^2\p B^2\Vp{\tfrac12}}}}}}{AB\Vp{\Vlr{\mb p}^3}}\p \texttt{constant} \end{split} \tl{B-06} \end{equation} QED.}

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^{(C) Interpretation of angle $\Theta$ in equation \eqref{10}}

Reference : My answer in What is the electric field flux through the base of a cube from a point charge infinitesimally close to a vertex?.

The angle $\Theta$ in equation \eqref{10} is a solid angle, not a plane one. This is explained in the above referenced answer of mine by equation (01) under the paragraph Proposition-Practical Rule. For completeness the Figure-01 and Figure-03 therein are copied as Figure-05 and Figure-06 respectively herein.

Above referenced equation (01) is proved in Section Differential Geometry. In this proving Section we encounter equations with integrals identical to these herein. For example the indefinite integral equations \eqref{dg-08} and \eqref{dg-11} therein

\begin{equation} \int\dfrac{1}{\bigl(x^{2}\!+\!y^{2}\!+\!R^{2}\bigr)^{\frac32}}\mathrm{d}x =\dfrac{x}{\left(y^{2}\!+\!R^{2}\right)\sqrt{y^{2}\!+\!x^{2}\!+\!R^{2}}}+\text{constant} \tl{dg-08} \end{equation}

\begin{equation} \int \dfrac{a}{\left(y^{2}\!+\!R^{2}\right)\sqrt{y^{2}\!+\!a^{2}\!+\!R^{2}}} \mathrm{d}y=\dfrac{\arctan\left(\dfrac{a y}{R\sqrt{y^{2}\!+\!a^{2}\!+\!R^{2}}}\right)}{R}+\text{constant} \tl{dg-11} \end{equation} are identical to \eqref{05} and \eqref{08} respectively herein.

Note that in the referenced answer and in equation \eqref{dg-07}
\begin{equation} \mathrm{S} =R^{3}\int\limits_{y=0}^{y=b}\int\limits_{x=0}^{x=a}\dfrac{1}{\bigl(x^{2}\!+\!y^{2}\!+\!R^{2}\bigr)^{\frac32}}\mathrm{d}x \mathrm{d}y \tl{dg-07} \end{equation} $\:\mr S\:$ is the surface area of a spherical patch, the (inverse stereographic) projection of rectangle $\:\mr{ADBC}\:$ on the sphere with center at point $\:Q\:$ and radius $\:R$, see Figure-06. \begin{equation} \Theta =\dfrac{\mathrm{S}}{R^{2}}=\arctan\left(\dfrac{a\,b}{R\sqrt{a^{2}\!+\!b^{2}\!+\!R^{2}}}\right)=\arctan\left(\dfrac{a\,b}{R\, d}\right) \tl{dg-13} \end{equation} is the solid angle by which the point $\:Q\:$ "sees" the rectangle $\:\mr{ADBC}$.

By analogy in equation \eqref{04} \begin{equation} \mr S \e h^3 \int\limits_{y\e \m b}^{y\e \p b}\blr{\:\:\int\limits_{x\e \m a}^{x\e \p a}\!\!\dfrac{\mr dx}{\plr{x^2\p y^2\p h^2}^{3/2}}}\mr dy \tl{C-01} \end{equation} is the surface area of a spherical patch, the (inverse stereographic) projection of rectangle $\:\texttt Q_1\texttt Q_2\texttt Q_3\texttt Q_4$ on the sphere with center at point $\:\mr P\:$ and radius $\:h$, see Figure-01, so \begin{equation} \Omega \e \dfrac{\mr S}{h^2}\e h \int\limits_{y\e \m b}^{y\e \p b}\blr{\:\:\int\limits_{x\e \m a}^{x\e \p a}\!\!\dfrac{\mr dx}{\plr{x^2\p y^2\p h^2}^{3/2}}}\mr dy \tl{C-02} \end{equation} is the solid angle by which the point $\:\texttt P\:$ "sees" the rectangle $\:\texttt Q_1\texttt Q_2\texttt Q_3\texttt Q_4$, see Figure-04.

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Answer 4

Actually this integral can be solved by the method of polar substitutions. x=rcos(A) and y=rsin(A) where r is the distance and A the angle in the polar plane. You can find further details in Thomas Calculus. Be sure to substitute the limits properly and multiply the integral by the Jacobian which in this case is r. Hope this answer helped you.

Answer 5

This integral cannot be solved in terms of elementary functions. You can easily do an expansion in $\frac{1}{r}$ in the integrand after doing on of the integrations, then doing the second integral after expanding you get $$ \frac{ab}{r^2}\left(1 - \frac{a^2+b^2}{12 r^2} + \mathcal{O}\left( \frac{1}{r^4}\right)\right) $$ If you want to solve the poisson equation, you have to use Green's function method because you have a charge distribution (unlike when you only have laplace equation with boundary conditions and you can just use separation of variables), this will bring you right back to this integral.

Note: this series is converging if you're interested in the region $r>\text{max }\left(a,b \right)$

Note: this is basically the multipole expansion, where the first term is the monopole contribution, the second is the quadrupole etc... (all odd multipole vanish because of symmetry)