Question

an orthorhombic box with all surfaces of this box are maintained at zero potential except the right surface at y=b. See picture I attached below. Find V=(x,y,z)!

I will give a medal and fan :)

Answer 1

Can help me @IGreen @dan815 @sleepyjess @Michele_Lalno @JFrase @Joel_the_boss @Inkyvoyd @rational ?

Answer 2

@Michele_Laino @JFraser

Answer 3

This is a genuinely three-dimensional problem\[\frac{ \partial^2 V }{ \partial x^2 }+\frac{ \partial^2 V }{ \partial y^2 }+\frac{ \partial^2 V }{ \partial z^2 }=0\]

Answer 4

subject to the boundary condition

Answer 5

(i) V=0 when x =0
(ii) V=0 when x=a
(iii) V=0 when z=0
(iv) V=0 when z=k
(v) V=0 when y=0
(vi) V=V(x,z) when y=b

Answer 6

We look for solutions that are products
\[V(x,y,z) = X(x)Y(y)Z(z)\]

Answer 7

Putting this into the first equation, and dividing by V, we find
\[\frac{ 1 }{ X } \frac{ \partial^2 X }{ dx^2 }+\frac{ 1 }{ Y } \frac{ \partial^2 Y }{ dy^2 }+\frac{ 1 }{ Z} \frac{ \partial^2 Z }{ dz^2 }=0\]

Answer 8

it follows that
\[\frac{ 1 }{ X } \frac{ \partial^2 X }{ dx^2 } =C_1\]
\[\frac{ 1 }{ Y } \frac{ \partial^2 Y }{ dy^2 } =C_2\]
\[\frac{ 1 }{ Z } \frac{ \partial^2 Z }{ dz^2 } =C_3\]

Answer 9

with \[C_1+C_2+C_3 =0\]

Answer 10

It suggests that \(C_1\) must be positive, \(C_2\) and \(C_3\) negative. Setting \(C_2 = -k^2\) and \(C_3 = -l^2\), we have \(C_1 = k^2+l^2\), and hence
\[\frac{ d^2X }{ dx^2 }=(k^2+l^2)X\]
\[\frac{ d^2Y }{ dy^2 }=(-k^2)Y\]
\[\frac{ d^2Z }{ dz^2 }=(-l^2)Z\]

Answer 11

Of course, separation of variable has turned a partial differential equation into ordinary differential equations. The solutions are

\[X(x)=A \sin(\sqrt{k^2+l^2}x)+B \cos(\sqrt{k^2+l^2}x)\]
\[Y(y)=C \sin(ky)+D \cos(ky)\]
\[Z(z)=E \sin(lz)+F \cos(lz)\]

Answer 12

Boundary condition (i) implies B=0, (iii) gives F=0, and (v) gives D=0, whereas (ii) and (iv) require that
\[\sqrt{K^2+l^2}=\frac{ n \pi }{ a }\]
\[l = \frac{ m \pi }{ k }\]
Thus,
\[K = \pi \sqrt{\frac{ n^2 }{ a^2 }-\frac{ m^2 }{ k^2 }}\]

where n and m are positive integers

Answer 13

@UnkleRhaukus

Answer 14

i am stuck :)

Answer 15

Help me
@perl @Hero @UnkleRhaukus @zepdrix @TheSmartOne @Nnesha @wio

Answer 16

@Nurali

Answer 17

can you help me @gorv ?

Answer 18

@Joel_the_boss

Answer 19

\(\frac{1}{Z}\frac{\partial^2 Z}{\partial z^2}\) harus menghasilkan fungsi Z(z) yg \(V(x,y,0)=X(x)Y(y)Z(0)=0\) dan \(V(x,y,k)=X(x)Y(y)Z(k)=U\)

Ambil aja \(\frac{1}{Z}\frac{\partial^2 Z}{\partial z^2}=\alpha^2, \frac{1}{Y}\frac{\partial^2 Y}{\partial y^2}=\beta^2\), dan \(\frac{1}{X}\frac{\partial^2 X}{\partial x^2}=-\gamma^2\) yg kemudian akan didapatkan \(X=X_0 \cos{\gamma x}\), \(Y=Y_0 \cos{\beta y}\), dan \(Z=Z_0 \sinh{\alpha z}\) dengan \(\alpha=\sqrt{\beta^2+\gamma^2}\). Dgn fungsi2 ini, \(X(0)\), \(Y(0)\), \(X(a)\), dan \(Y(b)\) gak nol, karena memang ga boleh nol agar saat \(Z(0)=0\) dan \(Z(k)\) gak nol.

Answer 20

@gorv

Answer 21

helllloooo please help @gerryliyana everybody!

Answer 22

@gerryliyana
can you please explain why the Laplacian is appropriate in this case?

It will take you case no further forward but i think it would be good to know.

Answer 23

I dont know @IrishBoy123