Electric Fields,Circuits
Electric Fields;
Emf,Currents,Potential
Difference and Multiloop Circuits
Lecture 21
Thursday: 1 April 2004
Coulomb’s Law
F 
1
40
1
q1 q2
40
r2
 8.99 109 Nm 2 /C 2
Direction is determined by opposites attract
and like charges repel one another.
Recall the Coulomb Force
Problem on Two Charges
q
qo
What is the force on q0 due to q if q= 2 C , qo = 3 C and the
distance between them is 3 meters?
How about if qo = 3 C ?
How about if qo = 1.5 C? …..
Fon q0 due to q 
1
q qo
40 r
2
Calculating the force exerted by q on any given charge you
might place at the location of qo gets repetitive. The only thing
that changes is the value of qo.
The Electric Field

Eq 

Fqqo
qo
• We define the electric field associated with a charge or
charge distribution to be the electrostatic force exerted per
unit of charge on which the force acts.
• If we know E, the force, F, on any charge, qo, is then
given by:
• We can calculate E once and then get F for any charge
easily.


Fqqo  qoEq


F
E
q0
The Electric Field
• From a point charge q, the force on q0 is,
F 
1
q q0
40 r
q
2
qo
• Then, at q0’s location, the field associated with q is
E 
F
q0

1
q
40 r 2
• Field is a vector. The direction is away from a
positive q and toward a negative q. It is also always in
the direction of F on a positive charge. 

F  qo E
Example
Example
T sin   qE
T cos  mg
T sin  qE

T cos mg
qE
tan  
mg