F) Coulomb's Law for two fixed point charges

r = distance between charges

F = k q1 q2
(N) r2

F -  mutual force of attraction or repulsion

k - electrostatic constant -

8.99 x 109 Nm2/C2
(p. 1 Reference)

q - charge of each object

 


 

Ex) What is the electrical force between 2 very small objects located .50 m apart when the charge on one object is 4.0 x 10-8 C and the charge on the second object is 6.0 x 10-5 C

 

 

Ex) What is the electrical force between 2 very small objects located .50 m apart when the charge on one object is 4.0 x 10-8 C and the charge on the second object is 6.0 x 10-5 C

 

r = .50 meters

q1 = 4.0 x 10-8 C

q2 = 6.0 x 10-5 C

k = 8.99 x 109 Nm2/C2
(p. 1 Reference)

 

F = k q1 q2
(N) r2

 

F = (8.99 x 109 Nm2/C2) (4.0 x 10-8 C) 6.0 x 10-5 C
(N) (.50 m)2

 

 

F = 8.6 x 10–2 Newtons

 


 

Relationships

F = k q1 q2
(N) r2

F,r

 

Inverse Square

 

Relationship between
F and r?

 

 

How would doubling the distance between the charges change the force between them?

 

 

F would be 1/4 as great

 

 F,q ?

F = k q1 q2
(N) r2

 

 

Direct

 



 

Two charges attract each other with a force of F.  If one charged was doubled and the other charge was tripled, how would that change the attractive force between these charges?

F = k q1 q2
(N) r2

 

 

Force increases 6 times