Contributed by:
Electric field, Electric field lines, Electric dipole, Field due to a dipole, Electric fields from continuous distributions
2.
The Electric Field
• Group of fixed charges exert a force F, given by
Coulomb’s law, on a test charge qtest at position r.
qtest
F
r
• The electric field E (at a given point in space) is the
force per unit charge that would be experienced by
a test charge at that point.
E = F / qtest This is a vector function of position.
3.
Electric Field of a Point Charge
F
qtest 1 Qq
test ˆ
F 2 r
r̂ r 4 0 r
• Dividing out qtest gives the electric field at r:
1 Q Radially outward,
E( r) ˆ
2 r falling off as 1/r2
4 0 r
4.
The Electric Field
Strength of electric field
Direction of the
electric field Superposition of electric field
5.
Electric Field Lines
Electric field lines (lines of force) are continuous lines
whose direction is everywhere that of the electric field
Electric field lines:
1) Point in the direction of the electric field E
2) Start at positive charges or at infinity
3) End at negative charges or at infinity
4) Are more dense where the field has greater magnitude
7.
Electric Field Lines (Point Charge)
Electric Field Field Lines
(vector) (Lines of
force)
Electric field lines (lines of force) are continuous lines
whose direction is everywhere that of the electric field
8.
Force Due to an Electric Field
F=qE
Just turn the definition of E around.
q
If E(r) is known, the force F on a
charge q, at point r is:
+
F = q E(r)
The electric field at r
Electric field lines are
points in the direction
bunched closer where
that a positive charge
the field is stronger.
placed at r would be
pushed.
9.
Two point charges, + 2 C each, are located on the x axis.
One charge is at x = 1 m, and the other at x = - 1 m.
a) Calculate the electric field at the origin.
b) Calculate (and plot) the electric field along the + y axis.
c) Calculate the force exerted on a + 5 C charge, located
at an arbitrary location on the + y axis
10.
The Electric Dipole
+q
d
-q
An electric dipole consists of two equal and opposite
charges (q and -q ) separated a distance d.
11.
The Electric Dipole
+q
d
p
-q
We define the Dipole Moment p
magnitude = qd,
p
direction = from -q to +q
12.
The Electric Dipole
E
+q
d
-q
Suppose the dipole is placed in a uniform electric
field (i.e., E is the same everywhere in space).
Will the dipole move ??
13.
The Electric Dipole
E
+q
d
-q
What is the total force acting on the dipole?
14.
The Electric Dipole
E F+
+q
d
F-
-q
What is the total force acting on the dipole?
15.
The Electric Dipole
E F+
+q
d
F-
-q
What is the total force acting on the dipole?
Zero, because the force on the two charges cancel:
both have magnitude qE. The center of mass does not
16.
The Electric Dipole
E F+
+q
d
F-
-q
What is the total force acting on the dipole?
Zero, because the force on the two charges cancel:
both have magnitude qE. The center of mass does not
But the charges start to move (rotate). Why?
17.
The Electric Dipole
E F+
+q
d
F-
-q
What is the total force acting on the dipole?
Zero, because the force on the two charges cancel:
both have magnitude qE. The center of mass does not
But the charges start to move (rotate). Why?
There’s a torque because the forces aren’t colinear.
18.
F+
d +q d sin
F-
-
q
The torque is:
magnitude of force) (moment arm)
(qE)(d sin
and the direction of is (in this case)
into the page
19.
d +q E
p
-q
but we have defined : p = q d
and the direction of p is from -q to +q
Then, the torque can be written as:
pxE = p E sin
with an associated potential energy
U = - p.E
20.
Electric fields due to various
charge distributions
The electric field is a vector which
obeys the superposition principle.
The electric field of a charge
distribution is the sum of the fields
produced by individual charges,
or by differential elements of charge
22.
Field Due to an Electric Dipole
at a point x straight out from its midpoint
Y
Electric dipole moment
p = qd
+q
l
d X
x
E- E+
-q
E
Calculate E as a function of p, x, and d, and approximate for x >> d
23.
Y
+q
l
d x X
E- E+
-q
E
24.
Electric Fields From Continuous
Distributions of Charge
Up to now we have only considered the electric field of point charges.
Now let’s look at continuous distributions of charge
lines - surfaces - volumes of charge
and determine the resulting electric fields.
Sphere Ring Sheet
25.
Electric Fields Produced by
Continuous Distributions of Charge
For discrete point charges, we can use the
Superposition Principle,
and sum the fields due to each point charge:
q2 Electric field
E experienced
q3 by q4
q1 q4
26.
Electric Fields From Continuous Distributions
For discrete point charges, we can use the superposition
principle and sum the fields due to each point charge:
q2
q3 E(r) Ei
i
q1 q4
What if we now have a continuous charge distribution?
q E(r)
27.
Electric Field Produced by a
Continuous Distribution of Charge
In the case of a continuous distribution of charge
we first divide the distribution up into small pieces,
and then we sum the contribution, to the field,
from each piece:
dEi
The small piece of charge dqi
r
produces a small field dEi at
the point r
Note: dqi and dEi are differentials
dqi
In the limit of very small pieces, the sum is an integral
28.
Electric Field Produced by a
Continuous Distribution of Charge
In the case of a continuous distribution of charge we first
divide the distribution up into small pieces, and then we
sum the contribution, to the field, from each piece:
In the limit of very small pieces, the sum is an integral
dq
Each dq: dE(r) k 2 r
dEi r
r Then: E = dEi
For very small pieces: E ( r ) dE
dqi
kdq
E( r ) 2 r
r
29.
Example: An infinite thin line of charge
Y
Find the electric
P field E at point P
y
X
Charge per unit length
is
30.
dE+
P
r x 2 y2
y
dq
x
dq
dE k r
r2 • Consider small element
of charge, dq, at position x.
• dq is distance r from P.
31.
dE
E yE
dE+ dE-
dq dq
-x x
dq
dE k r
r2 • Consider small element
of charge, dq, at position x.
dq
dE y 2k 2 cos • dq is distance r from P.
r
• For each dq at +x, there
is a dq at -x.
32.
dE
dE+ dE-
dq dq
-x x
dq
dE k r
r2
dq
dE y 2k 2 cos
r
dq= dx, cos=y/r
33.
dE
dE+ dE-
dq dq
-x x
dq
dE k r 2k dx y
r2 dE y 2 2
(x y ) r
dq
dE y 2k 2 cos x
r
2k y
E y 2 2 3/2 dx
dq= dx, cos=y/r
x 0
(x y )
34.
dE
dE+ dE-
dq dq
-x x
dq
dE k r 2k dx y
r2 dE y 2 2
(x y ) r
dq
dE y 2k 2 cos x
r 2k y 2k
E y 2 2 3/2 dx
dq= dx, cos=y/r x 0
(x y ) y
35.
Example of Continuous Distribution:
Ring of Charge
Find the electric field at a point along the axis.
Hint: be sure to use the symmetry of the problem!
dE
z r Break the charge up
into little bits, and find
the field due to each
dq bit at the observation
point. Then integrate.
Thin ring with total charge q R
charge per unit length is
q/R
36.
Continuous Charge Distributions
LINE AREA VOLUME
charge density =Q/L =Q/A =Q/V
units C/m C / m2 C / m3
differential dq = dL dq = dA dq = dV
Charge differential dq kdq
to be used when finding
the electric field of a
E( r ) 2 r
continuous charge distribution r
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