initial commit / add all books

19 files changed, 343 insertions, 0 deletions

diff --git a/1544/CH3/EX3.1/Ch03Ex1.sce b/1544/CH3/EX3.1/Ch03Ex1.sce
new file mode 100755
index 000000000..a39396408
--- /dev/null
+++ b/1544/CH3/EX3.1/Ch03Ex1.sce
@@ -0,0 +1,10 @@
+// Scilab code Ex3.1: Pg 79 (2008)

+clc; clear;

+Q = 50e-03;                 // Electric charge, C

+A = 600e-06;                // Area of plate, m^2

+// Solving for electric field density, D

+D = Q/A;                      // Electric field density, C/m^2

+printf("\nThe density of the electric field existing between the plates = %4.1f C/m-square", D);

+

+// Result

+// The density of the electric field existing between the plates = 83.3 C/m-square

diff --git a/1544/CH3/EX3.10/Ch03Ex10.sce b/1544/CH3/EX3.10/Ch03Ex10.sce
new file mode 100755
index 000000000..6ec4e12d5
--- /dev/null
+++ b/1544/CH3/EX3.10/Ch03Ex10.sce
@@ -0,0 +1,23 @@
+// Scilab code Ex3.10: Pg 90-91 (2008)

+clc; clear;

+C_1 = 6e-06;                    //Capacitance, F

+C_2 = 4e-06;                   //Capacitance, F

+V = 150;                      // Supply voltage, V

+// Part (a)

+// The reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal of the individual capacitances present in the circuit i.e 1/C =  1/C1 + 1/C2, solving for C

+C = ( C_1*C_2 )/(C_1 + C_2);               // Resulting capacitance, F

+// Part (b)

+Q = V*C;                                  // Electric charge on the capacitors, C

+// Part (c)

+V_1 = Q/C_1;                              // P.d across capacitor C_1, V

+V_2 = Q/C_2;                              // P.d across capacitor C_2, V

+printf("\nThe total capacitance of the combination = %3.1f micro-farad", C/1e-06);

+printf("\nThe charge on each capacitor = %3d micro-coulomb",Q/1e-06);

+printf("\nThe p.d. developed across %1d micro-farad capacitor = %2d V", C_1/1e-06, V_1);

+printf("\nThe p.d. developed across %1d micro-farad capacitor = %2d V", C_2/1e-06, V_2);

+

+// Result

+// The total capacitance of the combination = 2.4 micro-farad

+// The charge on each capacitor = 360 micro-coulomb

+// The p.d. developed across 6 micro-farad capacitor = 60 V

+// The p.d. developed across 4 micro-farad capacitor = 90 V

diff --git a/1544/CH3/EX3.11/Ch03Ex11.sce b/1544/CH3/EX3.11/Ch03Ex11.sce
new file mode 100755
index 000000000..079962202
--- /dev/null
+++ b/1544/CH3/EX3.11/Ch03Ex11.sce
@@ -0,0 +1,21 @@
+// Scilab code Ex3.11: Pg 91-92 (2008)

+clc; clear;

+C_1 = 3e-06;                    //Capacitance, F

+C_3 = 12e-06;                   //Capacitance, F

+C_2 = 6e-06;                    //Capacitance, F

+V = 400;                        // Supply voltage, V

+// The reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal of the individual capacitances present in the circuit i.e 1/C =  1/C1 + 1/C2 + 1/C_3, solving for C

+C = (C_1 * C_2 * C_3)/( C_1*C_2 + C_2*C_3 + C_3*C_1);               // Resulting capacitance, F

+Q = V*C;                                  // Electric charge on the capacitors, C

+// Part (c)

+V_1 = Q/C_1;                              // P.d across capacitor C_1, V

+V_2 = Q/C_2;                              // P.d across capacitor C_2, V

+V_3 =Q/C_3;                               // P.d across capacitor C_2, V

+printf("\nP.d across capacitor %1d micro-farad = %5.1f V", C_1/1e-06, V_1);

+printf("\nP.d across capacitor %1d micro-farad = %5.1f V", C_2/1e-06, V_2);

+printf("\nP.d across capacitor %2d micro-farad = %4.1f V", C_3/1e-06, V_3);

+

+// Result

+// P.d across capacitor 3 micro-farad = 228.6 V

+// P.d across capacitor 6 micro-farad = 114.3 V

+// P.d across capacitor 12 micro-farad = 57.1 V

diff --git a/1544/CH3/EX3.12/Ch03Ex12.sce b/1544/CH3/EX3.12/Ch03Ex12.sce
new file mode 100755
index 000000000..72ddda713
--- /dev/null
+++ b/1544/CH3/EX3.12/Ch03Ex12.sce
@@ -0,0 +1,43 @@
+// Scilab code Ex3.12: Pg 92-95 (2008)

+clc; clear;

+V = 200;                      // Supply voltage, voltage

+C_AB = 2;                // Capacitance across branch AB, micro-farad

+C_BC = 3;                 // Capacitance across branch BC, micro-farad

+C_CD = 6;                 // Capacitance across branch CD, micro-farad

+C_EF = 8;                // Capacitance across branch EF, micro-farad

+C_BD = 4;                 // Capacitance across branch EF, micro-farad

+

+// Part (a)

+// Since 3-micro-farad & 6-micro-farad  capacitors are in series & the reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal of the individual capacitances present in the circuit, therefore i.e 1/C =  1/C1 + 1/C2 

+C_BCD = ( C_BC*C_CD )/(C_BC+C_CD);         // Resulting capacitance across branch BCD, micro-farad

+//Since C_BCD & 4-micro-farad  capacitors are in parallel & the resulting capacitance of parallerly connected capacitors is the sum of the individual capacitance present in the circuit

+C_BD = C_BCD + C_BD;         // Resulting capacitance across branch BD, micro-farad

+//  Since 2-micro-farad & C_BD  capacitors are in series & the reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal of the individual capacitances present in the circuit, therefore, we have

+C_AD = (C_BD*C_AB)/(C_BD+C_AB);         // Resulting capacitance across branch AD, micro-farad

+//Since C_AD & C_EF  capacitors are in parallel & the resulting capacitance of parallerly connected capacitors is the sum of the individual capacitance present in the circuit

+C = C_AD + C_EF;         // Resulting capacitance of the circuit, micro-farad

+Q = V*C;                                  // Electric  charge drawn from the supply, C

+

+// Part (b)

+Q_EF = V*C_EF;                        // The charge on the 8 micro-farad capacitor, micro-coulomb

+

+// Part (c)

+Q_AD = Q - Q_EF;                        // The charge on the 4 micro-farad capacitor, C

+Q_BD = Q_AD;    // Charge in series combination of capacitors, micro-farad

+// Since Q = C*V, solving for V

+V_BD = Q_BD/C_BD;                    // The p.d. across the 4  F capacitor,V

+

+// Part(d)

+Q_BCD = V_BD*C_BCD;                  // Electric charge across branch BCD, C

+Q_BC = Q_BCD;                        // Electric charge, C

+V_BC = Q_BC/C_BC;                    // The p.d. across the 3 micro-farad capacitor

+printf("\nThe charge drawn from the supply = %3.1f mC", Q/1e+03);

+printf("\nThe charge on the %1d micro-farad capacitor = %3.1f mC", C_EF, Q_EF/1e+03);

+printf("\nThe p.d. across the %1d micro-farad capacitor= %2d V", C_BD, V_BD);

+printf("\nThe p.d. across the %1d micro-farad capacitor = %5.2f V", Q_BC, V_BC);

+

+// Result 

+// The charge drawn from the supply = 1.9 mC

+// The charge on the 8 micro-farad capacitor = 1.6 mC

+// The p.d. across the 6 micro-farad capacitor= 50 V

+// The p.d. across the 100 micro-farad capacitor = 33.33 V 

diff --git a/1544/CH3/EX3.13/Ch03Ex13.sce b/1544/CH3/EX3.13/Ch03Ex13.sce
new file mode 100755
index 000000000..f17c1e926
--- /dev/null
+++ b/1544/CH3/EX3.13/Ch03Ex13.sce
@@ -0,0 +1,13 @@
+// Scilab code Ex3.13: Pg 96 (2008)

+clc; clear;

+N = 20;                  // Number of plates in a capacitor

+A = 6400e-06;            // Cross - sectional area of plate, m^2

+d = 1.5e-03;             // Distance between plates, m

+epsilon_r = 6.4;         // Relative permittivity for mica

+epsilon_o = 8.854e-12;     // Relative permittivity for free space

+// Calculating the capacitance of the capacitor

+C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d;       // Capacitance, F

+printf("\n The capacitance of the capacitor = %3.1f nF", C/1e-09);

+

+// Result

+// The capacitance of the capacitor = 4.6 nF

diff --git a/1544/CH3/EX3.14/Ch03Ex14.sce b/1544/CH3/EX3.14/Ch03Ex14.sce
new file mode 100755
index 000000000..808b8c28e
--- /dev/null
+++ b/1544/CH3/EX3.14/Ch03Ex14.sce
@@ -0,0 +1,13 @@
+// Scilab code Ex3.14: Pg 96-97 (2008)

+clc; clear;

+N = 9;                      // Number of plates in a capacitor

+A = 1200e-06;               // Cross - sectional area of plate, m^2

+C = 3e-10;                  // Capacitance, F

+epsilon_r = 5;              // Relative permittivity for mica

+epsilon_o = 8.854e-12;      // Relative permittivity for free space

+// Using the formula of capacitance, C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d and solving for d, we have

+d = ((epsilon_o)*(epsilon_r)*A*(N-1))/C;        // Distance between plates, m

+printf("\nThe thickness of mica between parallel plates of a capacitor = %4.2f mm", d/1e-03);

+

+// Result

+// The thickness of mica between parallel plates of a capacitor = 1.42 mm 

diff --git a/1544/CH3/EX3.15/Ch03Ex15.sce b/1544/CH3/EX3.15/Ch03Ex15.sce
new file mode 100755
index 000000000..e370aea52
--- /dev/null
+++ b/1544/CH3/EX3.15/Ch03Ex15.sce
@@ -0,0 +1,14 @@
+// Scilab code Ex3.15: Pg 97 (2008)

+clc; clear;

+N = 11;                       // Number of plates in a capacitor

+r = 25e-03;                   // Radius of circular plate, m

+A = (%pi*r^2);                // Cross - sectional area of plate, m^2

+d = 5e-04;                    // Distance between plates, m

+epsilon_r = 1;                // Relative permittivity for air

+epsilon_o = 8.854e-12;        // Relative permittivity for free space

+// Calculating the capacitance of the capacitor

+C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d;       // Capacitance, F

+printf("\n The capacitance of the capacitor = %3.2f pF", C/1e-10);

+

+// Result

+// The capacitance of the capacitor = 3.48 pF

diff --git a/1544/CH3/EX3.16/Ch03Ex16.sce b/1544/CH3/EX3.16/Ch03Ex16.sce
new file mode 100755
index 000000000..924ccb073
--- /dev/null
+++ b/1544/CH3/EX3.16/Ch03Ex16.sce
@@ -0,0 +1,21 @@
+// Scilab code Ex3.16: Pg 99 (2008)

+clc; clear;

+C_1 = 3e-06;                                  // Capacitance, F

+C_2 = 6e-06;                                  // Capacitance, F

+V_1 = 250;                                   // Voltage across capacitor C_1, V

+// Since each capacitor will take charge according to its capacitance, so we have

+Q = C_1*V_1;                                 // Charge on first capacitor C_1, C

+W_1 = 0.5*C_1*(V_1^2);                       // Energy stored, J

+// When the two capacitors are connected in parallel the 3 micro-farad will share its charge with 6 micro-farad capacitor. Thus the total charge in the system will remain unchanged, but the total capacitance will now be different

+C = C_1 + C_2;                              // Total capacitance, F

+// Since Q = C*V, solving for V

+V = Q/C;                                    // Voltage across capacitor C_2, V

+W = 0.5*C*(V^2);                            // Total energy stored by the combination, J

+printf("\nThe charge and energy stored by %1d micro-F capcitor are %3.2f mC and %5.2f mJ respectively ", C_1/1e-06, Q/1e-03 , W_1/1e-03);

+printf("\nThe p.d. between the plates = %5.2f V", V);

+printf("\nThe energy stored by the combination of %1d micro-F and %1d micro-F capacitors = %5.2f mJ", C_1/1e-06, C_2/1e-06, W/1e-03);

+

+// Result

+// The charge and energy stored by 3 micro-F capcitor are 0.75 mC and 93.75 mJ respectively 

+// The p.d. between the plates = 83.33 V

+// The energy stored by the combination of 3 micro-F and 6 micro-F capacitors = 31.25 mJ 

diff --git a/1544/CH3/EX3.17/Ch03Ex17.sce b/1544/CH3/EX3.17/Ch03Ex17.sce
new file mode 100755
index 000000000..29bad772a
--- /dev/null
+++ b/1544/CH3/EX3.17/Ch03Ex17.sce
@@ -0,0 +1,27 @@
+// Scilab code Ex3.17: Pg 99-100 (2008)

+clc; clear;

+V = 200;                       // Supply voltage, V

+C_1 = 10e-06;                  // Capacitance, farad

+C_2 = 6.8e-06;                 // Capacitance, farad

+C_3 = 4.7e-06;                 // Capacitance, farad

+// Part (a)

+// Since each capacitor will take charge according to its capacitance, so we have

+Q_1 = V*C_1;                   // Charge sored on  capacitor C_1, C

+W_1 = 0.5*C_1*(V^2);           // Energy sored on  capacitor C_1, J

+// Part (b)

+// Since C_2 and C_3 are in series and hence, their equivalent capacitance is given by their series combination

+C_4 = (C_2 * C_3)/(C_2 + C_3);          // Equivalent capacitance of C_2 and C_3, F

+// Since C_1 and C_4 are in parallel and hence, their equivalent capacitance is given by their parallel combination

+C = C_1 + C_4;               // Total capacitance of circuit, F

+// Since Q = C*V, solving for V

+V_1 = Q_1/C;                         // New p.d across C_1, V

+W = 0.5*C*(V_1^2);                 // Total energy remaining in the circuit, J

+energy_used = W_1 - W;             // Energy, J

+printf("\nThe charge and energy stored by %2d micro-F capacitor are %1d mC and %2.1f J respectively ", C_1/1e-06, Q_1/1e-03, W_1);

+printf("\nThe new p.d across %2d micro-F capacitor = %5.1f V", C_1/1e-06, V_1);

+printf("\nThe amount of energy used in charging %3.1f micro-F and %3.2f micro-F capacitors from %2d micro-F capacitor = %4.3f J", C_2/1e-06, C_3/1e-06, C_1/1e-06, energy_used/1e-03);

+ 

+// Result

+// The charge and energy stored by 10 micro-F capacitor are 2 mC and 0.2 J respectively 

+// The new p.d across 10 micro-F capacitor = 156.5 V

+// The amount of energy used in charging 6.8 micro-F and 4.70 micro-F capacitors from 10 micro-F capacitor = 43.495 J 

diff --git a/1544/CH3/EX3.18/Ch03Ex18.sce b/1544/CH3/EX3.18/Ch03Ex18.sce
new file mode 100755
index 000000000..f163f7241
--- /dev/null
+++ b/1544/CH3/EX3.18/Ch03Ex18.sce
@@ -0,0 +1,10 @@
+// Scilab code Ex3.18: Pg 101 (2008)

+clc; clear;

+V = 400;               // Supply voltage, V

+E = 0.5e06;            // Dielectric strength, V/m

+// Since E = V/d, solving for d

+d = V/E;               // Thickness of dielectric, m

+printf("\nThe minimum thickness of dielectric required = %3.1fmm", d/1e-03);

+

+// Result

+// The minimum thickness of dielectric required = 0.8 mm

diff --git a/1544/CH3/EX3.19/Ch03Ex19.sce b/1544/CH3/EX3.19/Ch03Ex19.sce
new file mode 100755
index 000000000..3f97143bd
--- /dev/null
+++ b/1544/CH3/EX3.19/Ch03Ex19.sce
@@ -0,0 +1,19 @@
+// Scilab code Ex3.19: Pg 101-102 (2008)

+clc; clear;

+C = 270e-12;              // Capacitance, F

+A = 60e-04;               // Cross-sectional area of plate, m^2

+E = 350e03;               // Dielectric strength, V/m

+epsilon_r = 2.1;           // Relative pemittivity

+epsilon_o = 8.854e-12;     // Permittivity of free space

+// Part (a)

+// Since formula for capacitance, C = ((epsilon_o)*(eplison_r)*A)/d, solving for d

+d = ((epsilon_o)*(epsilon_r)*A)/C;      // Thickness of dielectric, m

+// Part (b)

+// Since E = V/d, solving for V

+V = E*d;                                  // Maximum possible working voltage, V

+printf("\nThe thickness of Teflon sheet required = %5.4f mm", d/1e-03);

+printf("\nThe maximum possible working voltage for the capacitor = %5.1f V", V);

+

+// Result

+// The thickness of Teflon sheet required = 0.413 mm

+// The maximum possible working voltage for the capacitor = 144.6 V

diff --git a/1544/CH3/EX3.2/Ch03Ex2.sce b/1544/CH3/EX3.2/Ch03Ex2.sce
new file mode 100755
index 000000000..f72926e3e
--- /dev/null
+++ b/1544/CH3/EX3.2/Ch03Ex2.sce
@@ -0,0 +1,15 @@
+// Scilab code Ex3.2: Pg 80 (2008)

+clc; clear;

+A = 400e-06;              // Cross-sectional area of plate, m^2

+I = 50e-06;              // Source current, A

+t = 3;                   // Flow time of current, s

+// Since electric current is the rate of flow of charge i.e I = Q/t, solving for Q 

+Q = I*t;                 // Amount of charge on plates, C

+//Solving for density of the electric field between the plates

+D = Q/A;                    // Electric field density, C/m^2

+printf("\The charge on the plates = %3d micro-coloumb", Q/1e-06);

+printf("\nThe density of the electric field between the plates = %5.3f C/m-square", D);

+

+// Result

+// The charge on the plates = 150 micro-coloumb

+// The density of the electric field between the plates =0.375 C/m-square

diff --git a/1544/CH3/EX3.3/Ch03Ex3.sce b/1544/CH3/EX3.3/Ch03Ex3.sce
new file mode 100755
index 000000000..0ad86c1f9
--- /dev/null
+++ b/1544/CH3/EX3.3/Ch03Ex3.sce
@@ -0,0 +1,18 @@
+// Scilab code Ex3.3: Pg 83 (2008)

+clc; clear;

+d = 3e-03;                  // Thickness of dielectric, m

+Q = 35e-03;                 // Electric charge on plates, C

+V = 150;                   // Supply voltage, V

+A = 144e-06;              // Cross-sectional area of plates, m^2

+// Part (a)

+// Since electric field strength(E) = potential gradient therefore we have

+E = V/d;                    // Electric field strength, V/m

+// Part (b)

+// Solving for electric field density, D

+D = Q/A;                      // Electric field density, C/m^2

+printf("\nThe electric field strength = %2d kV/m", E*1e-03);

+printf("\nThe flux density = %5.1f C/m^2", D);

+

+// Result

+// The electric field strength = 50 kV/m

+// The flux density = 243.1 C/m^2

diff --git a/1544/CH3/EX3.4/Ch03Ex4.sce b/1544/CH3/EX3.4/Ch03Ex4.sce
new file mode 100755
index 000000000..4c4573845
--- /dev/null
+++ b/1544/CH3/EX3.4/Ch03Ex4.sce
@@ -0,0 +1,23 @@
+// Scilab code Ex3.4: Pg 83-84 (2008)

+clc; clear;

+d = 4e-03;                   // Thickness of air, m

+Q = 2e-04;                  // Electric charge on plates, C

+V = 125;                   // Supply voltage, V

+D = 15;                   // Electric field density, coulomb-per-metre-square

+// Part (a)

+// Since electric field strength(E) = potential gradient, therefore we have 

+E = V/d;                    // Electric field strength, V/m

+// Part (b)

+// Since D = Q/A, solving for A

+A = Q/D;              // Cross-sectional area of plates, m^2

+// Part (c)

+// Since Q = C*V, solving for C

+C = Q/V;                    // Capacitance of the plates, F

+printf("\nThe electric field strength between the plates = %5.2f kV/m",E*1e-03);

+printf("\nThe csa of the field between the plates = %4.1f mm^2", A/1e-06);

+printf("\nThe capacitance of the plates = %3.1f micro-coulomb", C/1e-06);

+

+// Result

+// The electric field strength between the plates = 31.25 kV/m

+// The csa of the field between the plates = 13.3 mm^2

+// The capacitance of the plates = 1.6 micro-coulomb

diff --git a/1544/CH3/EX3.5/Ch03Ex5.sce b/1544/CH3/EX3.5/Ch03Ex5.sce
new file mode 100755
index 000000000..d7830dfaf
--- /dev/null
+++ b/1544/CH3/EX3.5/Ch03Ex5.sce
@@ -0,0 +1,19 @@
+// Scilab code Ex3.5: Pg 86 (2008)

+clc; clear;

+A = 6e-04;                          // Cross-sectional area of plates, m^2

+d = 5e-04;                          // Thickness of mica sheet, m

+Epsilon_r = 5.8;                    // Relative permittivity, unitless

+Epsilon_0 = 8.854e-12;              // Permittivity of Free Space

+V = 200;                            // Potential difference, V

+// Part (a)

+// Since absolute permittivity, Epsilon = C*(d/A), therefore solving for d & putting Epsilon = Epsilon_0*Epsilon_r 

+C = ( Epsilon_r*Epsilon_0*A )/d;         // Capacitance, F

+// Part (b)

+// Since electric field strength(E) = potential gradient, therefore we have 

+E = V/d;                    // Electric field strength, V/m

+printf("\nThe capacitance of the capacitor = %5.2f pF", C/1e-12);

+printf("\nElectric field strength = %3d kV/m",E*1e-03);

+

+// Result

+// The capacitance of the capacitor = 61.62 pF

+// Electric field strength = 400 kV/m

diff --git a/1544/CH3/EX3.6/Ch03Ex6.sce b/1544/CH3/EX3.6/Ch03Ex6.sce
new file mode 100755
index 000000000..85ab7ecf2
--- /dev/null
+++ b/1544/CH3/EX3.6/Ch03Ex6.sce
@@ -0,0 +1,12 @@
+// Scilab code Ex3.6: Pg 86 (2008)

+clc; clear;

+C = 0.224e-09;                    //Capacitance, F

+A = 5625e-06;                     // Cross-sectional area of plates, m^2

+Epsilon_r = 2.5;                  // Relative permittivity

+Epsilon_0 = 8.854e-12;            // Permittivity of Free Space

+// Since absolute permittivity, Epsilon = C*(d/A), therefore solving for d & putting Epsilon = Epsilon_0*Epsilon_r

+d = ( Epsilon_r*Epsilon_0*A )/C;                  // Thickness of waxed paper dielectric, m

+printf("\nThe thickness of paper required = %3.2f mm", d/1e-03);

+

+// Result

+// The thickness of paper required = 0.56 mm

diff --git a/1544/CH3/EX3.7/Ch03Ex7.sce b/1544/CH3/EX3.7/Ch03Ex7.sce
new file mode 100755
index 000000000..4786ae59a
--- /dev/null
+++ b/1544/CH3/EX3.7/Ch03Ex7.sce
@@ -0,0 +1,12 @@
+// Scilab code Ex3.7: Pg 86 (2008)

+clc; clear;

+C = 4.7e-08;                   //Capacitance, F

+A = 4e-04;                   // Cross-sectional area of plates, m^2

+d = 1e-04;                  // Thickness of dielectric, m

+Epsilon_0 = 8.854e-12;            // Permittivity of Free Space

+// Since absolute permittivity, Epsilon = C*(d/A), therefore solving for Epsilon_r & putting Epsilon = Epsilon_0*Epsilon_r

+Epsilon_r = (C*d)/(Epsilon_0*A);                  // Relative permittivity

+printf("\nRelative permittivity = %4d", Epsilon_r);

+

+// Result

+// Relative permittivity = 1327

diff --git a/1544/CH3/EX3.8/Ch03Ex8.sce b/1544/CH3/EX3.8/Ch03Ex8.sce
new file mode 100755
index 000000000..990ee1d08
--- /dev/null
+++ b/1544/CH3/EX3.8/Ch03Ex8.sce
@@ -0,0 +1,19 @@
+// Scilab code Ex3.8: Pg 87 (2008)

+clc; clear;

+V = 180;                            // Potential difference, V

+d = 3e-03;                         // Thickness of dielectric, m

+A = 4.2e-04;                       // Cross-sectional area of plates, m^2

+Epsilon_r = 3.5;                   // Relative permittivity

+Epsilon_0 = 8.854e-12;             // Permittivity of Free Space

+// Since absolute permittivity, Epsilon = C*(d/A), therefore solving for C & putting Epsilon = Epsilon_0*Epsilon_r 

+C = ( Epsilon_r*Epsilon_0*A )/d;         // Capacitance, F

+// Since C = Q/V, solving for Q

+Q = C*V;                                 // Electric charge, C

+// Using  D = Q/A,

+ D = Q/A;                                // Electric field density, C/m^2

+ printf("\The flux thus produced = %3.2f nC.",Q/1e-09);

+ printf("\nThe flux density thus produced. = %3.2f micro-coulomb-per-metre-square", D/1e-06); 

+ 

+// Result

+// The flux thus produced = 0.78 nC

+// The flux density thus produced. = 1.86 micro-C/m^2

diff --git a/1544/CH3/EX3.9/Ch03Ex9.sce b/1544/CH3/EX3.9/Ch03Ex9.sce
new file mode 100755
index 000000000..9a5f2c889
--- /dev/null
+++ b/1544/CH3/EX3.9/Ch03Ex9.sce
@@ -0,0 +1,11 @@
+// Scilab code Ex3.9: Pg 89 (2008)

+clc; clear;

+C_1 = 4.7e-06;                   //Capacitance, F

+C_2 = 3.9e-06;                   //Capacitance, F

+C_3 = 2.2e-06;                   //Capacitance, F

+// The resulting capacitance of parallerly connected capacitors is the sum of the individual capacitance present in the circuit

+C = C_1 + C_2 + C_3;             // Resulting capacitance  of the circuit, F

+printf("\nThe resulting capacitance of the combination = %4.1f micro-farad", C/1e-06);

+

+// Result

+// The resulting capacitance of the combination = 10.8 micro-farad