There is a long direct metal rectangular slot, length a=40, width b=20, the potential of the side wall and bottom surface are zero, the potential of the top cover is 100V (relative value), and the potential distribution in the slot is obtained.
The solution code using Gaussian iteration is as follows (the maximum allowable error of two adjacent iteration values is 0.001):
a=zeros(21,41);
a(1,:)=100;
b=zeros(19,39);
c=eye(19,39);
count=1;
d=0;
while(count==1)
m=0;
for i=2:1:20
for j=2:1:40
b(i-1,j-1)=a(i,j);
a(i,j)=0.25*(a(i-1,j)+a(i+1,j)+a(i,j-1)+a(i,j+1));
c(i-1,j-1)=abs(a(i,j)-b(i-1,j-1));
if(c(i-1,j-1)<0.001)
for k=1:1:19
for n=1:1:39
if(c(k,n)<0.00001)
m=m+1;
else
m=0;
break;
end
end
if(m==0)
break;
end
if(m==741)
count=0;
end
end
end
if(count==0)
break;
end
end
if(count==0)
break;
end
end
d=d+1;
end
d
a
The procedure for using the over-relaxation method is as follows:
d=zeros(1,10);
h=0;
for e=1:0.1:1.9
a=zeros(21,41);
a(1,:)=100;
b=zeros(19,39);
c=eye(19,39);
count=1;
g=0;
while(count==1)
m=0;
for i=2:1:20
for j=2:1:40
b(i-1,j-1)=a(i,j);
a(i,j)=a(i,j)+e*0.25*(a(i-1,j)+a(i+1,j)+a(i,j-1)+a(i,j+1)-4*a(i,j));
c(i-1,j-1)=abs(a(i,j)-b(i-1,j-1));
if(c(i-1,j-1)<0.001)
for k=1:1:19
for n=1:1:39
if(c(k,n)<0.001)
m=m+1;
else
m=0;
break;
end
end
if(m==0)
break;
end
if(m==741)
count=0;
end
end
end
if(count==0)
break;
end
end
if(count==0)
break;
end
end
g=g+1;
end
h=h+1;
e
d(1,h)=g;
end
d
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