11. It states that the force of attraction or repulsion between
any two stationary point charges is directly proportional to
the product of their magnitudes of the two charges and
inversely proportional to the square of the distance
between them.
21. p=qL = the electric dipole moment
The dipole moment p is defined as a vector
directed
from -ve to +ve.
22. 22
θ−=+ cosE2EE x2x1
iˆcosE2E,So net θ−=
2
a
kq
E =
a2
L
a
cos 2
L
==θ
iˆ
a2
L
a
kq
2E,So 2net −=
r
Enet = −
kqL
a3
ˆi
2
2
2
L
ra
+=
( )( )
2
3
2
3 2
r2
L32
2
L2
net
)(1(
1
r
p
k
r
p
kE
+
=
+
=
qLp =
r
Enet
;
kp
r3
Electric Dipole
a
a
E
1
E1x
θ
θ
ˆi
ˆj
y
For large r
E1x = −E cosθ and E1x = E2x
E2
23. 23
A uniform external electric field exerts no net force on a dipole, but it
does exert torque that tends to rotate the dipole in the direction of the
field (align with )p
extE
1F
2F
x
Torque about the com = τ
= FLsinθ
= qELsinθ = pEsinθ =
r
p ×
r
E
Ep
×=τSo,
When the dipole rotates through θ, the electric field does work:
24. 24
Potential Energy
Ans. The energy is minimum when aligns with
EpcospEU
⋅−=θ−==
Integrating
,
So, U=-
p⋅
E
p
E
dW=−τdθ=−pEsinθdθWork done equals
The minus sign arises because the torque opposes any increase in θ.
Setting the negative of this work equal to the change in the potential
energy, we have
θθ+=−= dsinpEdWdU
0UcospEdsinpEdWdUU +θ−=θθ−=−== ∫ ∫ ∫
°=θ= 90when0UchooseWe
25. The electric flux through a given area held
inside an electric field is the measure of the
total number of electric lines of force
passing normally through that area
ΔΦ=E Δs cosθ
It is a scalar quantity and it’s SI unit is
(Nm*m)/c
It has direction which is normal to the plane.
26. States the total electric flux
through a closed surface
(surface integral of electric
field over a closed surface)
is equal to 1/εo times the
total charge enclosed by the
surface.
Mathematically
( )enclosed
s
qSdE∫ =
0
1
.
ε