### Electrostatics Formula :

Coulomb’s law of Electrostatics force F_{(v/m)} in volts per meter is equal to the coulomb’s constant k_{(Nm2c2) } in Newton per meter2 per coulomb and multiply the magnetic force of charges q_{1,} q_{2(c) } in coulomb and dived by square of the distance between two charges r^{2}_{(c)} in coulomb. Hence the coulomb’s law can be written as

**F=1/4πε _{0 }*qQ/r^{2} =k_{e} qQ /r^{2}**

^{or}

**F _{(v/m)}= k_{(Nm2c2) } * q_{1(c)} *q_{2(c)} /r^{2}_{(c)}**

where,

F→ magnetic force in volts per meter

K→ coulomb’s constant K=8.988*10^{9} Nm^{2 } c^{-2}

q_{1,} q_{2}→ magnetic force of charge in coulomb

r→ distance between two charges in centimeter

### Example:1

Calculate exert a force 18*10^{-3}N on each other what is the magnitude of each charges? Two equal and like charges are placed at a distance d =8cm.

### Answer:

F_{(v/m)}= k_{(Nm2c2) } * q_{1(c)} *q_{2(c)} /r^{2}_{(c)}

K= is coulomb’s constant and has a value 8.99*10^{9}Nm^{2}/c^{2}

Let the magnitude of charges be [q_{1}]= [q_{2}]= [q]

18*10^{-3}=(8.99*10^{9})*q^{2}/(0.08)^{2}

18*10^{-3}=(8.99*10^{9})*q^{2}/(0.0064)

q^{2}=2528*10^{-16}

q=1264*10^{-8} c

### Example:2

Calculate exert a force 12*10^{-3}N on each other what is the magnitude of each charges? Two equal and like charges are placed at a distance d =6cm.

### Answer:

F_{(v/m)}= k_{(Nm2c2) } * q_{1(c)} *q_{2(c)} /r^{2}_{(c)}

K= is coulomb’s constant and has a value 8.99*10^{9}Nm^{2}/c^{2}

Let the magnitude of charges be [q_{1}]= [q_{2}]= [q]

12*10^{-3}=(8.99*10^{9})*q^{2}/(0.06)^{2}

12*10^{-3}=(8.99*10^{9})*q^{2}/(0.0036)

q^{2}=48*10^{-16}

q=6.9*10^{-8} c