1. Electric Forces and Fields Electric Force

2. Coulomb’s Law Electric force – the force of attraction or repulsion between objects due to charge –Electric force makes solids feel solid –Electric force holds atoms together Like all forces it has units of Newtons (N) Electric force depends on charge and distance –The closer two charges are the greater the force between them

3. Coulomb’s Law Coulomb’s law relates electric force, charge, and distance –Named in honor of Charles Coulomb –Uses a constant called Coulomb’s constant (k C ) Electric force = coulomb’s constant * (charge 1) * (charge 2) / (distance)2 F electric = k C q 1 q 2 /r 2 k C has a value of 8.9875*10 9 Nm 2 /C 2 –We will use 8.99*10 9 Nm 2 /C 2 for our calculations

4. Coulomb’s Law Electric force is a field force like gravitational force The equations are very similar However while gravitational force is always attractive, electric force is either attractive or repulsive Electric force is much stronger –A small electric force can overcome the gravitational force

5. Coulomb’s Law Often more than two charges are present Find the individual charges using Coulomb’s law Use the Superposition principle – use vector addition (right triangle functions and Pythagorean theorem) to find net electric force

6. Coulomb’s Law

7. Equilibrium position – the location at which the net electric force on a charge is zero –Find charges using Coulomb’s law and set as equal –Solve for the distance between one charge and the equilibrium position

8. Coulomb’s Law Three charges lie along the x-axis. One positive charge, q 1 = 15μC, is at x=2.0m, and another positive charge, q 2 =6.0μC, is at the origin. At what point on the x-axis must a negative charge, q 3, be placed so that the resultant force on it is zero? Given: q 1 = 15μC (15*10 -6 C)r 3,1 = 2.0-P q 2 =6.0μC (6.0*10 -6 C)r 3,2 = P

9. Coulomb’s Law q 3 must be in the middle between q 1 and q 2 Set the F electric between q 1 and q 3 and between q 2 and q 3 equal to one another Use Coulomb’s law to solve for P F 3,1 =F 3,2 k C (q 3 q 1 /(r 3,1 ) 2 ) = k C (q 3 q 2 /(r 3,2 ) 2 ) k C (q 3 (15*10 -6 )/(2.0-P) 2 ) = k C (q 3 6.0*10 -6 /(P) 2 ) 15/(2.0-P) 2 ) = 6.0/ /(P) 2 ) P 2 (15) = (2.0-P) 2 (6.0) P=.80m from q 2

10. Coulomb’s Law Coulomb was the first person to quantify electric force and established the inverse square law for electric charges –Hence, Coulomb’s law, Coulomb’s constant, and the unit Coulomb Performed his experiments with a torsion balance of small spheres fixed on the ends of a rod –One sphere was given a charge and brought near another charged object –This caused the rod to rotate The faster the rotation, the greater the charge

11. Coulomb’s Law