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+//chapter 7
+//example 7.3
+//Calculate the radius of electron cloud and dispalcement
+//page 188
+clear;
+clc;
+//given
+N=2.7E25; // in 1/m^3 (density of atoms)
+E=1E6; // in V/m (electric field)
+Z=2; // atomic number of Helium 
+Eo=8.85E-12; // in F/m (absolute permittivity)
+Er=1.0000684; // (dielectric constant of the material)
+e=1.6E-19; // in C (charge of electron)
+pi=3.14; // value of pi used in the solution
+//calculate
+// since  alpha=Eo*(Er-1)/N=4*pi*Eo*r_0^3 
+// Therefore we have r_0^3=(Er-1)/(4*pi*N)
+r_0=((Er-1)/(4*pi*N))^(1/3);// calculation of radius of electron cloud
+printf('\nThe radius of electron cloud is \t r_0=%1.2E m',r_0);
+x=4*pi*Eo*E*r_0/(Z*e); // calculation of dispalcement
+printf('\n\nThe displacement is x=%1.2E m',x);
+// NOTE: The answer is wrong due to calculation mistake.