Electrostatics Formula and Sample Problems

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 q1, q2(c)  in coulomb and dived by  square of the distance between two charges r2(c) in coulomb. Hence the coulomb’s law can be written as

F=1/4πε*qQ/r2 =ke qQ /r2

or

F(v/m)= k(Nm2c2)  * q1(c) *q2(c) /r2(c)

where,

F→ magnetic force in volts per meter

K→ coulomb’s constant K=8.988*109 Nm2  c-2

q1, q2→ magnetic force of charge in coulomb

r→ distance between two charges in centimeter

Example:1

Calculate exert a force 18*10-3N 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)  * q1(c) *q2(c) /r2(c)

K= is coulomb’s constant and has a value 8.99*109Nm2/c2

Let the magnitude of charges be [q1]= [q2]= [q]

18*10-3=(8.99*109)*q2/(0.08)2

18*10-3=(8.99*109)*q2/(0.0064)

q2=2528*10-16

q=1264*10-8 c

Example:2

Calculate exert a force 12*10-3N 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)  * q1(c) *q2(c) /r2(c)

K= is coulomb’s constant and has a value 8.99*109Nm2/c2

Let the magnitude of charges be [q1]= [q2]= [q]

12*10-3=(8.99*109)*q2/(0.06)2

12*10-3=(8.99*109)*q2/(0.0036)

q2=48*10-16

q=6.9*10-8 c

Leave a Comment