Transcript Electric Fields - E. R. Greenman

Electric Fields
PHYSICS CHAPTER 22 A
Electric Field
 Definition:
 Michael Faraday (1791-1867)
 Electric field exists around any charged object
 Area where a test charge would be given a force by the charged
object
 Test charge is a positive charge of small magnitude
 Magnitude of field is a measure of the force exerted by the
charged object
 Area of field indicated by field lines
Show the path that would be taken by the test charge
 Note: electric field intensity is a VECTOR!

Electric Field


Electric field is said to exist in
the region of space around a
charged object: the source
charge.
Concept of test charge:



Small and positive
Does not affect charge
distribution
Electric field intensity:



 F
E
q0
+ +
+ +
+
+
+
+ +
Existence of an electric field is
a property of its source;
Presence of test charge is not
necessary for the field to exist;
September 18, 2007
The electric field will be directly proportional to the charge setting up the
field and inversely proportional to the square of the distance between that
charge and where you are measuring the strength of the electric field.
In other words,
Electric field = (Coulomb’s constant) (Charge on object setting up the field)
(distance away from the charge)2
E is in N/C
E = kq
Q2
is in Coulombs
d
K=9x109 Nm2
C2
Distance is in
meters
Example of Field Lines
 Electric Field Lines
Problem p 370
 A positive test charge of 4.0 x 10-5 C is placed in an
electric field. The force acting on the test charge is
0.60 N at 10o. What is the electric field intensity at
the location of the test charge?
 USE THE STEPS!
 What we know:


Q = 4.0 x 10-5C
F = 0.60 N at 10o
 Equation:
 E = F/q
Problem continued
 Substitute in values:
 E = 0.60 N

4.0 x 10-5C
 Solve the math:
 E = 1.5 x 104 N/C at 10o
 Check answer!
 Reasonable answer?
 Units???
Problem : What is the electric field 0.2 meters away from a +4
Coulomb charge?
Equation: E = kq/d2
E = (9x109 Nm2/C2)(4C)
(0.2 m)2
The direction of the electric field would be away from
the +4C charge since the direction is always the direction
a small positive test charge would move.
E=9x1011 N/C