Transcript Electric field due to a dipole

Physics
ELECTROSTATICS - 2
Session Objectives
Electric field due to a point charge
Electric field due to a dipole
Torque and potential energy of a dipole
Electric lines of force
Gauss’ Law
Electric field due to a point charge
The test charge q0 is placed at M. The force on it,

1 qq0 
F=
d
3
4ε0 d
Electric field strength at M,
E=
Or,
F
1
q
=
d
3
q0
4ε0 d
E=
1 q
4ε0 d2
The direction of electric filed is away
from the positive point charge
Solved Example - 1
Find the electric field due to a point charge of 0.5
mC at a distance of 4 cm from it in vacuum. How
does the filed strength change if the charge is in
a medium of relative permittivity 80?
Solution:
q =0.5×10–3 C, d = 4×10–2 m, k = 80
1 q
0.5×10–3
9
10
–1
(i) E=
=9×10
×
=2.82×10
NC
2
4 ε0 d2
 4×10–2 
1
q 2.82×1010
8
–1
(ii) E'=
=
=3.53×10
NC
4 ε0k d2
80
Superposition Principle
Electric field due to a number of charges at a point is
the vector sum of the fields due to individual charges.
Rules of vector addition, as is the case for
electric forces, is to be applied.
Solved Example –2
Two point charges +q and +4q are separated by a
distance 6a. Find the point on the line joining the
charges where the electric field is zero.
Solution:
The field due to the charges should be equal
and opposite at this point. Let the point be at
a distance x from the charge +q. Then,
1 q
1
4q
=
 x = 2a
4ε0 x2 4ε0 (6a– x)2
The point is at a distance 2a
from the charge +q
+q
+4q
x
6a–x
Electric Dipole
A system of two equal and opposite
charges separated by a certain distance
Dipole moment - a measure of the
strength of electric dipole. It is a vector
quantity represented by P .
Magnitude of dipole moment - product of the magnitude of
either charge and the separation between them.
Direction of dipole moment – it points from
negative towards positive charge
P = q (2a)
SI unit – C m
Solved Example - 3
Two charges 20 mC and – 20 mC are 4 mm
apart. Determine the electric dipole moment of
the system. If the charges are in a medium,
will the dipole moment change?
Solution:
q = 20 ×10–6 C,
2a = 4×10–3 m
(i)
p = q(2a) = 20×10–6×4×10–3 = 8 ×10–8 C m
(ii)
Dipole moment is independent of the nature of
the medium between the charges
Electric field due to a dipole
(i) Along the axial line
The resultant field at M,



E = E1+E2

E=
q
2
4ε0  d+ a
ˆ
q .n
=
4ε0

E=
ˆ
×(–n)+


 2d(2a) 

2
2
2
 d -a




2p
4ε0 d3


q
2
4ε0  d– a
×nˆ

=
2p
3
4ε0 d
ˆ
;as p =q(2a) n
and d >> a
Electric field due to a dipole
(ii) On the equatorial plan

E1 =

E2 =
q
4ε0 (d2 + a2 )
q
4ε0 (d2 + a2 )
E
Sine components of E1 and E2
cancel out. Cos components add up

E = E1 cosθ+E2 cosθ
G

E =
=
2q
2
2
4ε0 (d + a )
cosθ
q×2a


2 3/2
4ε0 d2 + a
as cosθ =
a
d2 + a2

p
=
3
4ε0d
as p = q(2a) and d>> a
The direction of electric field is opposite to that of dipole moment
Therefore, E =
p
4ε0 d3
Torque on a dipole
In an electric field E, the force acting on charges will be equal
and opposite. The parallel forces form a couple. Moment of
the couple constitute a torque
t=
=
=
=
force x perpendicular distance
qE x AC
qE x 2a sinq
PE sinq
– qE



τ = p× E
E
A
qE
q
B
;where, p is the dipole moment
C
Solved Example – 4
An electric dipole, when held at 30° with respect to
a uniform electric field of 104 NC–1, experiences a
torque of 9 ×10–26 Nm. Find its dipole moment.
Solution:
t = 9×10–26 Nm,
q =30°,
E =104 NC–1
t = PEsinθ
τ
9×10–26
P=
= 4
=1.8×10–29 Cm
Esinθ 10 ×sin30°
Potential Energy of a dipole
Work done on a dipole in an electric field is stored as
its potential energy
Work done in rotating a dipole from
orientation q1 to q2
W


q2
q1

q2
q1
t(q) dq
pE sin q dq
q
 –pE cos qq2
1
=-pEcosθ2 - cosθ1 
If q1 = 90° and q2 = q, then,
W = -pE cosθ- cos90o  = -pEcosθ


Or,
Potential energy of the dipole,
U=–pE cosθ =–p.E
Solved Example – 5
What does the negative sign in the expression for
the potential energy of a dipole indicate? ( U=–p.E )
Solution:
The potential energy of the dipole is taken to be zero,
When it is oriented perpendicular to the direction of the
electric field. It is the maximum potential energy, the
dipole can acquire. In any other orientation, it is less than
zero and hence, the negative sign.
Electric line of force
Imaginary curve drawn in an electric field
whose tangent at any point gives the the
direction of intensity of the field at that point.
Lines of force originate at a positive charge
and terminate at a negative charge.
Tangent at any point to line of force shows
the direction of the field at that point.
Lines of force do not intersect and are
continuous curves.
Number density of lines of force at point
gives the magnitude of electric field.
Electric Field lines
Point Charge
+ve : radial, outward
directed st. lines
end at 
-ve : radial, inward
directed st. lines
begin at 
Both are uniformly distributed.
No. of lines for charge |q| = q/
Electric Field lines
2. Two Positive Charges
M is the neutral point
Equal charges: same number density
When the charges are unequal, M shifts
towards the smaller charge.
Electric Field lines
3. Two unlike Charges
Unequal charges
Equal charges:
same number density
–Ve > + Ve
Electric Field lines
Uniform Field
From infinity
Equispaced parallel straight lines
To infinity
Solved Example - 6
Why cannot two electric lines of force
intersect each other?
Solution:
The tangent at a point on the line of force gives the
direction of the field at that point. If two lines of force
intersect at a point, the field will have two directions at
that point, which is impossible.
Electric Flux
Electric flux over a surface is equivalent to the
number of electric field lines passing normally
through it. Denoted by f
Mathematically, f   E.ds
Integration is carried over the surface and ds is a small
area element. The direction of ds is perpendicular to
the surface element
SI unit of f : N m2 C–1.
Gauss’ Law
It relates electric fields at points on
a closed surface and the net charge
enclosed by that surface.
The flux over a closed surface is
1
times charge enclosed by that surface
ε0
qenc
f=
ε0
But
ds
f =  E.ds
s
Therefore, ε0
 E.ds = qenc
s
qenc
+ +
- + + -
q
E
Gauss’ Law
If no charge is enclosed by the
surface, fnet = 0
f
q
in
f
out
Gauss law is useful in computing electric fields
if the charge distribution is symmetric
Solved Example - 7
The net outward flux through the surface of a black
box is 8 ×103 Nm2C–1.
(a) What is the net charge inside the box?
(b) If the net outward flux were zero, could you say
that there were no charges inside the box?
Solution:
(a) q = 8 ×103 Nm2C–1
If the net charge inside is q, then,
q=
q
 q = Φ ε0 = 8×103 ×8.854×10–12 = 7.08×10–8 C
ε0
(b) Not necessarily. We can only say that the net
charge inside the box is zero.
Thank You