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Transcript
Electric Field
Physics Department, New York
City College of Technology
Key words






Test charge
Electric force
Electric field
Point charge
Superposition
principle
Electric field lines
Test charge

What is test Charge?

A charge so small
that the force it
exerts does not
significantly alter the
distribution of other
charges that are
nearby.
Definition of electric field


The electric field, E , at
any point in space is
defined as the force
exerted on a tiny
positive test charge
placed at that point
divided by the
magnitude of the test
charge q:

 F
E
q
Electric field


The electric field is a
vector whose direction
is the direction of the
force on a tiny positive
test charge at that
point, and whose
magnitude is the force
per unit charge.
The SI units of E : N/C
Electric field

Disc 17, #10
Electric field of a point charge

The electric field at a distance r from a
single point charge Q has magnitude
qQ
k 2
F
E  r
q
q

, therefore, E  k
Q
r2
E is independent of the test charge q,
that is, E depends only on the charge Q
which produces the field
Example #1

Find vector E for (a) and (b).
Example #1--Continued
(9.0 109 N  m2 / C 2 )(3.0 106 C )
5
Ek 2 

3
.
0

10
N /m
2
r
(0.30m)
Q
The direction is the electric field is toward the charge Q.
Calculate electric force


If E is given, then



If q is positive, F and E



F  qE
point in the same
direction.


If q is negative, F and E
point in opposite
directions.

F
Superposition principle

The electric field due to more than one
charge is the vector sum of all the
individual fields due to each charge:
 

E  E1  E 2  ...
Example #2

Calculate E at P.
Example #2--Continued
Ek
Q1
2
1
r
k
Q2
r22
25 10 6 C
50 10 6 C
 (9.0 10 N  m / C )(

)
2
2
2
2
(2.0 10 m) (8.0 10 m)
9
2
2
 6.3 108 N / C
The electric field points to the left.
Example #3
Example #3--Continued
(9.0 109 N  m 2 / C 2 )(50 10 6 C )
6
E A1 

1
.
25

10
N /C
2
(0.60m)
E A2
(9.0 109 N  m 2 / C 2 )(50 10 6 C )
6


5
.
0

10
N /C
2
(0.30m)
E Ax  E A1 cos 330  E A2 cos 90  1.1106 N / C
E Ay  E A1 sin 330  E A 2 sin 90  4.4 106 N / C
E A  1.12  4.4 2 106 N / C
tan  
E Ay
E Ax
 4.0,   76
Example #3--Continued
EB1  EB 2
(9.0 109 N  m 2 / C 2 )(50 10 6 C )
k 2 
r
(0.4m) 2
Q
 2.8 106 N / C
EB  2EB1 cos   2(2.8 106 N / C )(0.65)  3.6 106 N / C
The direction of EB is along the +x direction.
Electric field lines

A pictorial description of the electric
field
Electric field lines



They indicate the direction of the electric field; the
field points in the direction tangent to the field line
at any point.
The lines are drawn so that the magnitude of the
electric field, E, is proportional to the number of
lines crossing unit area perpendicular to the lines.
The closer together the lines, the stronger the field.
They start on positive charges and end on negative
charges; and the number starting or ending is
proportional to the magnitude of the charge.
Electric field lines
Electric fields and conductors



The electric field inside a
conductor is zero in the
static situation.
Any net charge on a
conductor distributes itself
on the surface.
The electric field is always
perpendicular to the
surface outside of a
conductor.