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//clear//
//Caption: Program to find the work involved 'W' in moving a charge 'Q' along shorter arc of a circle
//Example4.1
//page 84
clc;
x = sym('x');
y = sym('y');
z = sym('z');
y1 = sym('y1');
y = sqrt(1-x^2);
Q = 2; //charge in coulombs
Edot_dL1 = integ(y,x);
disp(Edot_dL1,'E.dx*ax =')
Edot_dL1 = limit(Edot_dL1,x,0.8)-limit(Edot_dL1,x,1);
disp(Edot_dL1,'Value of E.dx*ax =')
Edot_dL2 = 0;
disp(Edot_dL2,'Value of E.dz*az=')
x = sqrt(1-y1^2);
Edot_dL3 = integ(x,y1)
disp(Edot_dL3,'E.dy*ay=')
Edot_dL3 = limit(Edot_dL3,y1,0.6)-limit(Edot_dL3,y1,0);
disp(Edot_dL3,'Value of E.dy*ay =')
W = -Q*(Edot_dL1+Edot_dL2+Edot_dL3);
disp(W,'Work done in moving a point charge along shorter arc of circle in Joules, W=')
//Result
// E.dx*ax = asin(x)/2+x*sqrt(1-x^2)/2
// Value of E.dx*ax = (25*asin(4/5)+12)/50-%pi/4
// Value of E.dz*az = 0.
// E.dy*ay = asin(y1)/2+y1*sqrt(1-y1^2)/2
// Value of E.dy*ay = (25*asin(3/5)+12)/50
//Work done in moving a point charge along shorter arc of circle in Joules, W =
// -2*((25*asin(4/5)+12)/50+(25*asin(3/5)+12)/50-%pi/4)
//Which is equivalent to
// -2*((25*0.9272952+12)/50+(25*0.6435011+12)/50-%pi/4) = -0.96 Joules
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