Transcript Coulomb`s Law

Prepared By
Waqas Amin Sheikh
 He
used a torsion balance
 For point charges ,charged bodies that are
very small in comparison with the distance r
between them
 He found that electric force is proportional
to 1/r2
 When
the distance is doubles , the force
decreases to ¼ of its initial value
 When
the distance halved the force increases
to four(4) times to its initial value
 The force between two point charges also
depends upon the quantity of charge on each
body
 To
explore this dependence he….
 Divided a charge into two equal parts by
placing a small charged spherical conductor
into contact with an identical but uncharged
sphere
 By symmetry the charge is shared equally
between the two spheres
 He
found that forces that the two point
charges q1 and q2 exert on each other are
proportional to each charges and therefore
are proportional to the product q1q2 of the
two charges
 The
magnitude of the electric force between
two point charges is directly proportional to
the product of the charges inversely
proportional to the square of the distance
between them
 If
q1 and q2 are the two point charges
separated by a distance r between them.
 Then according to coulomb’s law
 F α q1.q2
 F α 1/r2
 F = k|q1.q2|/r2
 Where k is the constant of proportionality
and its value is depend upon the system of
unit used
 The
absolute values bars are used in
equation because the charges q1 and q2 can
be either positive or negative
 While the force F will always remain positive
 When the both charges q1 and q2 have the
same sign either positive or negative the
forces are repulsive
 When the both charges q1 and q2 have the
opposite sign positive and negative the
forces are attractive
 The
value of the constant k is depends upon
the system of unit used
 In our study of electromagnetism we will use
SI unit there fore the value of k
 K=8.987551787 * 109N.m2/C2
 K=1/4πe0
 1/4πe0=9.0*109 N.m2/C2
 one coulomb =6*1018electrons
 Experiments
shows that when two charges
exerts a force simultaneously on a third
charge, the total force acting on that charge
is the vector sum of the forces that the two
charges would exert individually
 When
two electrically charged particles in
empty space interact, how does each one
know the other is there?
 What goes on in the space between them to
communicate the effect of each one to the
other?
 To introduce this concept look at the fig.




Look at the repulsive
forces of two positively
charged bodies
Let B has the charge q0
Let F0 be the electric
force of A on B
This is an force of
“action at distance”
that acts across empty
space without needing
any matter i.e( such as
a push rod or rope) to
transmit it through the
intervening space
 We
first envision
that body A ,as a
result of the charge
that it carries
,somehow modifies
the properties of
the space around it.
 Then body B as a
result of the charge
that it carries,
sense how space has
been modified at its
position
 The
response of
body B is to
experience the
force F0
 To
elaborate how
this two stage
process occur we
first consider body A
by itself we remove
body B and label its
former position as
point P
 Charged
body A
produces or causes
an electric field at
point P. and all
other points in the
neighborhood.
 This electric field is
present at P even if
there is no other
charge at P
 This is due to the
charge of body A.
If a point charge qo is
then placed at point P
it experience a force
Fo
 We take a point of
view that the force
exerted on qo by the
field at P
 Thus electric field is
the intermediary
through which A
communicates its
presence to qo

 The
point charge
qo would
experience a force
at any point in the
neighborhood of A
 The electric field
that A produces
exists at all points
in the region
around A
 We
can say that the
point charge qo
produces an electric
field in the space
around it and this
electric field exerts
the force – Fo on
body A
 For each forces
(force on A due to
qo and on qo due to
A)
 One
charge sets up
an electric field
that exerts a force
on the second
charge
 This is an
interaction between
two charges
 A single charge
created an electric
field in the
surroundings space ,
but this electric
field cannot exert a
 The
electric force
on a charged body
is exerted by the
electric field
created by the
other charges
bodies
 To
find out there
is an electric field
we place a
charged body
which we call a
test charge at the
point(shown in fig.
c)
 If the test charge
experience an
electric force then
there is as electric
field at the point
 Electric
field E :
the electric field E
at a point as the
electric force Fo
experienced by a
test charge qo .
Thus the electric
field at a certain
point is equal to
the electric force
per unit charge
experienced by a

E = Fo/ qo
The SI unit in which
the unit of force is 1
N and the unit of
charge is 1C the unit
of electric field
magnitude is 1
newton per coulomb
 If the field E is known
at any point the
electric force is given
by
 Fo = E* qo


Qo
+
Fo
E


Qo
Fo
-
E
The charge qo can be
+ve or can be –ve .
If qo is +ve the force
will be in the same
direction as E and vice
versa
caution : the electric
force experienced by a
test charge qo can
vary from point to
point so the electric
field can also be
different at different
points
 Since
the electric field E can vary from point
to point it is not a single vector quantity but
an infinite set of vectors quantities
 We call the location of the charge source
point and the point P where we determine
the field is called field point
 It
is also useful to introduce a unit vector r
that points along the line from source to the
field point
 This unit vector is equal to the displacement
vector r from source to the field point
divided by the distance r between these two
points i.e
 r =r/r
 if we place a small test charge qo at the
field point P at a distance r from the source
point the magnitude Fo of the force is given
by the coulomb’s law
 Fo = 1/4πE
|qqo|/r2
 Magnitude of E of the electric field at P is
 E = 1/4πe |q|.r/r2
 using the unit vector r . We can write a
vector equation that gives both magnitude
and direction of the electric field
 E = 1/4πe q.r/r2
 Using
equation below
 E = 1/4πe q.r/r2
 we can measure the electric field caused by
a single point charge
 But in most cases the electric field is
distributed over space
 The charged plastic and glass rod have
electric charge distributed over their
surfaces
 Field
caused by charge distribution we can
imaElectric field calculation
 gine the distribution to be made up of many
charges q1, q2, q3,………
 At any point P each point charge produces
its own electric field E 1, E 2, E 3,….
 So a test charge qo placed at P experience a
force F1= q0E 1 from charge q1 a force
f2=q0E2
 From charge q2 and so on
 The
total force F0 = F1 + F2 +F3……..= q0E 1+ q0E
2….
combine effect of all the charges in the
distribution is describe by the total electric
field E at point P . From the def. of electric
field
 E = Fo/ qo = E1 + E2 + E3……..
 The total electric field at P is the vector sum
of the field at P due to each point charge in
the charge distribution this is the “ Principle
of the superposition of electric field”
 The