An analytical solution for the potential field due to a uniformly charged rectangular plate has been examined in the following two recent articles :
1. D . A . Fagundes, "Electrostatic potential of a rectangular uniformly charged plate: exact solution and limiting cases," Euro . J . Phys. 43, 015203 (2021).
2. O. Ciftja and B. Ciftja, “Exact analytical results for the electrostatic potential due to a uniformly charged finite plate,” Can. J. Phys. 101, 607-6012 (2023).
In Ref. [2] , the potential V(x,y,z) for a finite rectangular plate lying on the x-y plane has been brought to the one-dimensional integral expression in Eq. (10). Only limiting results for V(0,0,z) are derived explicitly. Using a different approach, the same limiting potential V(0,0,z) has been obtained in Ref [1].
​
It is the goal in this note to show that using Mathematica one can obtain the potential field and the corresponding electric field at an arbitrary point in space.​
​
The potential field due to a uniformly charged rectangular plate of dimensions 2a and 2b is given as follows:
V(x,y,z)=
σ
4πϵ
b
∫
-b
a
∫
-a
y'x'
1
2
(x-x')
+
2
(y-y')
+
2
z
;
σ is the constant charge surface on the plate and ϵ denotes the permittivity .
The change of variables
x - x' = X, y - y' = Y
results in the expression
V(x,y,z)=
σ
4πϵ
y+b
∫
y-b
x+a
∫
x-a
YX
1
2
X
+
2
Y
+
2
z
.
The integration over X is performed first .
V(x,y,z)=
σ
4πϵ
x+a
∫
x-a
YX
1
2
X
+
2
Y
+
2
z
Integrate
1
2
X
+
2
Y
+
2
z
,{X,x-a,x+a}
-Log-a+x+
2
Y
+
2
(a-x)
+
2
z
+Loga+x+
2
Y
+
2
(a+x)
+
2
z
 if
condition
This expression is rewritten a follows :
Log
a+x+
2
Y
+
2
(a+x)
+
2
z
-a+x+
2
Y
+
2
(a-x)
+
2
z
;
The integration over Y is performed next .
IntegrateLog
a+x+
2
Y
+
2
(a+x)
+
2
z
-a+x+
2
Y
+
2
(a-x)
+
2
z
,{Y,y-b,y+b}
zArcTan
(a-x)(-b+y)
z
2
(a-x)
+
2
(-b+y)
+
2
z
+zArcTan
(a+x)(-b+y)
z
2
(a+x)
+
2
(-b+y)
+
2
z
-zArcTan
(a-x)(b+y)
z
2
(a-x)
+
2
(b+y)
+
2
z
-zArcTan
(a+x)(b+y)
z
2
(a+x)
+
2
(b+y)
+
2
z
-bLog-a+x+
2
a
+
2
b
-2ax+
2
x
-2by+
2
y
+
2
z
+yLog-a+x+
2
a
+
2
b
-2ax+
2
x
-2by+
2
y
+
2
z
-aLog-b+y+
2
a
+
2
b
-2ax+
2
x
-2by+
2
y
+
2
z
+xLog-b+y+
2
a
+
2
b
-2ax+
2
x
-2by+
2
y
+
2
z
+bLoga+x+
2
(a+x)
+
2
(-b+y)
+
2
z
-yLoga+x+
2
(a+x)
+
2
(-b+y)
+
2
z
-aLog-b+y+
2
(a+x)
+
2
(-b+y)
+
2
z
-xLog-b+y+
2
(a+x)
+
2
(-b+y)
+
2
z
-bLog-a+x+
2
(a-x)
+
2
(b+y)
+
2
z
-yLog-a+x+
2
(a-x)
+
2
(b+y)
+
2
z
+aLogb+y+
2
(a-x)
+
2
(b+y)
+
2
z
-xLogb+y+
2
(a-x)
+
2
(b+y)
+
2
z
+bLoga+x+
2
(a+x)
+
2
(b+y)
+
2
z
+yLoga+x+
2
(a+x)
+
2
(b+y)
+
2
z
+aLogb+y+
2
(a+x)
+
2
(b+y)
+
2
z
+xLogb+y+
2
(a+x)
+
2
(b+y)
+
2
z
 if
condition
This solution is simplified as follows:
V=
σ
4πϵ
-zArcTan
(a-x)(b-y)
z
2
(a-x)
+
2
(b-y)
+
2
z
+ArcTan
(a+x)(b-y)
z
2
(a+x)
+
2
(b-y)
+
2
z
+ArcTan
(a-x)(b+y)
z
2
(a-x)
+
2
(b+y)
+
2
z
+ArcTan
(a+x)(b+y)
z
2
(a+x)
+
2
(b+y)
+
2
z
+(b-y)Log
a+x+
2
(a+x)
+
2
(b-y)
+
2
z
-a+x+
2
(a-x)
+
2
(b-y)
+
2
z
+(a-x)Log
b+y+
2
(a-x)
+
2
(b+y)
+
2
z
-b+y+
2
(a-x)
+
2
(b-y)
+
2
z
-(b+y)Log
-a+x+
2
(a-x)
+
2
(b+y)
+
2
z
a+x+
2
(a+x)
+
2
(b+y)
+
2
z
+(a+x)Log
b+y+
2
(a+x)
+
2
(b+y)
+
2
z
-b+y+
2
(a+x)
+
2
(b-y)
+
2
z
;
This is the potential field V (x, y, z) at an arbitrary point in space .
Along the axis of symmetry one has
V/.x->0/.y->0//FullSimplify​​
1
4πϵ
σ-4zArcTan
ab
z
2
a
+
2
b
+
2
z
-bLog
-a+
2
a
+
2
b
+
2
z
a+
2
a
+
2
b
+
2
z
+bLog
a+
2
a
+
2
b
+
2
z
-a+
2
a
+
2
b
+
2
z
+2aLog
b+
2
a
+
2
b
+
2
z
-b+
2
a
+
2
b
+
2
z

or, more simply,
V0=
1
4πϵ
σ-4zArcTan
ab
z
2
a
+
2
b
+
2
z
+bLog
2
a+
2
a
+
2
b
+
2
z

2
-a+
2
a
+
2
b
+
2
z

+2aLog
b+
2
a
+
2
b
+
2
z
-b+
2
a
+
2
b
+
2
z
;
This expression coincides with those given in the Refs [1] and [2].
R12=
σ
4πϵ
4aArcSinh
b
z^2+a^2
+bArcSinh
a
z^2+b^2
-zArcTan
ab
z
z^2+a^2+b^2
;
V0/.{a->1,b->2,z->2}//N
0.271953σ
ϵ
R12/.{a->1,b->2,z->2}//N
0.271953σ
ϵ
The exact electric field at an arbitrary point is given by
e=-Grad[V,{x,y,z}];
e1=e
σ
4πϵ
/.y->0/.a->1/.b->2;
Below,vectorplotsoftheelectricfeldareshownonthey=0plane.
e1=e
σ
4πϵ
/.y->0/.a->1/.b->2;
VectorPlot[{e1.{1,0,0},e1.{0,0,1}},{x,-3,3},{z,-3,3},PlotRange->{1/16,3}]
VectorPlot[{e1.{1,0,0},e1.{0,0,1}},{x,-3,3},{z,-3,3},PlotRange->{-3,-1/16}]
The electric field along the axis of symmetry is given as follows :
e0=-
σ
4πϵ
Grad[Sol,{x,y,z}]/.{x->0,y->0}//FullSimplify
0,
σLog
-a+
2
a
+
2
b
+
2
z
a+
2
a
+
2
b
+
2
z
+Log
a+
2
a
+
2
b
+
2
z
-a+
2
a
+
2
b
+
2
z

4πϵ
,
σArcTan
ab
z
2
a
+
2
b
+
2
z

πϵ

The z - component is identical to that given in Eq. (5) of Ref. [1] obtained from -dV(0,0,z)/dz.

CITE THIS NOTEBOOK

Potential field of a charged rectangular plate​
​by Ioannis M. Besieris​
Wolfram Community, STAFF PICKS, November 4, 2025
​https://community.wolfram.com/groups/-/m/t/3570273