Filter löschen
Filter löschen

it is being difficult plotting the graph for Capacitance vs Distance D..please provide me a suitable solution to plot the graph

1 Ansicht (letzte 30 Tage)
%USING THE METHOD OF MOMENT,
% THIS PROGRAM DETERMINES THE CAPACITANCE OF A
% PARALLEL-PLATE CAPACITOR CONSISTING OF TWO CONDUCTING
% PLATES, EACH OF DIMENSION AA x BB, SEPARATED BY A
% DISTANCE D, AND MAINTAINED AT 1 VOLT AND -1 VOLT
% ONE PLATE IS LOCATED ON THE Z = 0 PLANE WHILE THE OTHER
% IS LOCATED ON THE Z=D PLANE
% ALL DIMENSIONS ARE IN S.I. UNITS
% N IS THE NUMBER IS SUBSECTIONS INTO WHICH EACH PLATE IS
%DIVIDED
% FIRST, SPECIFY THE PARAMETERS
ER = 1.0;
EO = 8.8541e-12;
AA = 1.0;
BB = 1.0;
D = 1.0;
N = 9;
NT = 2*N;
M = sqrt(N);
DX = AA/M;
DY = BB/M;
DL = DX;
% SECOND, CALCULATE THE ELEMENTS OF THE COEFFICIENT
% MATRIX A
K = 0;
for Kl=1:2
for K2=1:M
for K3=1:M
K = K + 1;
X(K) = DX*(K2 - 0.5);
Y(K) = DY*(K3 - 0.5);
end
end
end
for K1=1:N
Z(K1) = 0.0;
Z(K1+N) = D;
end
for I=1:NT
for J=1:NT
if(I==J)
A(I,J) = DL*0.8814/(pi*EO);
else
R = sqrt( (X(I)-X(J))^2 + ( Y(I)-Y(J) )^2 + ( Z(I)-Z(J) )^2 ) ;
A(I,J) = DL^2/(4.*pi*EO*R);
end
end
end
% NOW DETERMINE THE MATRIX OF CONSTANT VECTOR B
for K=1:N
B(K) = 1.0;
B(K+N) = -1.0;
end
% INVERT A AND CALCULATE RHO CONSISTING
% THE UNKNOWN ELEMENTS
% ALSO CALCULATE THE TOTAL CHARGE Q AND CAPACITANCE C
F = inv(A);
RHO = F* (B');
SUM = 0.0;
for I=1:N
SUM = SUM + RHO(I);
end
Q = SUM*(DL^2) ;
VO = 2.0;
C = abs(Q)/VO;
Please provide a suitable solution to plot the graph Capacitance vs Distance
  6 Kommentare

Melden Sie sich an, um zu kommentieren.

Antworten (1)

Youssef  Khmou
Youssef Khmou am 2 Feb. 2014
Bearbeitet: Youssef Khmou am 2 Feb. 2014
you have to measure each time the capacitance with varying d, there is peak at d=0.1, is this reasonable ? :
%USING THE METHOD OF MOMENT,
% THIS PROGRAM DETERMINES THE CAPACITANCE OF A
% PARALLEL-PLATE CAPACITOR CONSISTING OF TWO CONDUCTING
% PLATES, EACH OF DIMENSION AA x BB, SEPARATED BY A
% DISTANCE D, AND MAINTAINED AT 1 VOLT AND -1 VOLT
% ONE PLATE IS LOCATED ON THE Z = 0 PLANE WHILE THE OTHER
% IS LOCATED ON THE Z=D PLANE
% ALL DIMENSIONS ARE IN S.I. UNITS
% N IS THE NUMBER IS SUBSECTIONS INTO WHICH EACH PLATE IS
%DIVIDED
% FIRST, SPECIFY THE PARAMETERS
ER = 1.0;
EO = 8.8541e-12;
AA = 1.0;
BB = 1.0;
N = 9;
NT = 2*N;
M = sqrt(N);
DX = AA/M;
DY = BB/M;
DL = DX;
D=0:0.001:1;
for n=1:length(D);
% SECOND, CALCULATE THE ELEMENTS OF THE COEFFICIENT
% MATRIX A
K = 0;
for Kl=1:2
for K2=1:M
for K3=1:M
K = K + 1;
X(K) = DX*(K2 - 0.5);
Y(K) = DY*(K3 - 0.5);
end
end
end
for K1=1:N
Z(K1) = 0.0;
Z(K1+N) = D(n);
end
for I=1:NT
for J=1:NT
if(I==J)
A(I,J) = DL*0.8814/(pi*EO);
else
R = sqrt( (X(I)-X(J))^2 + ( Y(I)-Y(J) )^2 + ( Z(I)-Z(J) )^2 ) ;
A(I,J) = DL^2/(4.*pi*EO*R);
end
end
end
% NOW DETERMINE THE MATRIX OF CONSTANT VECTOR B
for K=1:N
B(K) = 1.0;
B(K+N) = -1.0;
end
% INVERT A AND CALCULATE RHO CONSISTING
% THE UNKNOWN ELEMENTS
% ALSO CALCULATE THE TOTAL CHARGE Q AND CAPACITANCE C
F = inv(A);
RHO = F* (B');
SUM = 0.0;
for I=1:N
SUM = SUM + RHO(I);
end
Q = SUM*(DL^2) ;
VO = 2.0;
C(n) = abs(Q)/VO;
end
figure, plot(D,C);
  6 Kommentare
ASUTOSH
ASUTOSH am 3 Feb. 2014
any help on this ...because if i change N its showing error as mentioned above..so please help me...
Patrik Ek
Patrik Ek am 3 Feb. 2014
Bearbeitet: Patrik Ek am 3 Feb. 2014
check the loop
for Kl=1:2
for K2=1:M
for K3=1:M
K = K + 1;
X(K) = DX*(K2 - 0.5);
Y(K) = DY*(K3 - 0.5);
end
end
end
The variables X and Y seems to be dependent on sqrt(N), which will cause trouble with N = 20. It would be more stable to give M a value and define N as
N = M^2;

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu 2-D and 3-D Plots finden Sie in Help Center und File Exchange

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by