# CHAPTER04 : THE FORCES¶

## Example E1 : Pg 59¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 4.1
# calculation of coulomb force

# given data
np=26.# number of protops in an iron atom
na=6.*10.**26.# number of atome in 58 kg iron
mi=58.# mass(in kg) of iron
e=1.6*10.**(-19.)# charge(in coulomb) on an electron
perdiff=1.# percentage of charge of electron less than that of proton
r=1.# separation(in m) between the two blocks

# calculation
poschrg=(na*np*e*perdiff)/(mi*100.)
fc=(9.*10.**9.*poschrg*poschrg)/(r*r)

print'the coulomb force(in newton) between the two blocks is newton',fc

the coulomb force(in newton) between the two blocks is newton 1.66677003567e+21


# WORKED EXAMPLES¶

## Example E3w : Pg 61¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 4.3w
# calculation of ratio of the electrical force to the gravitational force between two electrons

# given data
me=9.1*10.**-31.# mass(in kg)of an electron
e=1.6*10.**-19.# charge(in coulomb)of an electron
k=9.*10.**9.# value of ratio 1/(4*%pi*epsilonzero)   (in N m**2/C**2)
G=6.67*10.**-11.# value of universal gravitational constant   (in N m**2/kg**2)

# calculation
ratio=(k*e*e)/(G*me*me)# ratio = electric force / gravitational force

print'the ratio of electric to gravitational force between two electrons is',ratio

the ratio of electric to gravitational force between two electrons is 4.1713233469e+42