calculating note for field intensity on equatorial line of electric dipole
Answers
Answer:
Therefore the Electric field due to an electric dipole is 4.1 × 10^5 N/C.
Given :
An electric dipole of length 10cm having charge of one micro coulomb at an equatorial point 12cm from the centre of the dipole.
To find :
The electric field due to the electric dipole.
solution :
The electric field due to an electric dipole is given by, E = KP/(r² + a²)³/²
where P = q2a = 10^-6 C × 10 × 10^-2 m = 10^-7 Cm
r = 12 cm = 0.12 m and a = 5cm = 0.05 m
The electric field, E = (9 × 10^9 × 10^-7)/[(0.12)² + (0.05)²]³/²
= (900)/[0.0144 + 0.0025]³/²
= 900/(0.0169)³/²
= 900/(0.13)³
= 900/0.002197
= 409,649.522
≈ 4.1 × 10^5 N/C
Therefore the Electric field due to an electric dipole is 4.1 × 10^5 N/C.
Answer:
Consider an electric dipole AB. The charges −q and +q of dipole are situated at A and B respectively as shown in the figure. The separation between the charges is 2a.
The direction of dipole moment is from −q to +q.
(i) At axial or end-on position : Consider a point P on the axis of dipole at a distance r from mid-point O of electric dipole.
The distance of point P from charge +q at B is
BP=r−a
And distance of point P from charge −q at A is, AP=r+a.
Let E
1
and E
2
be the electric field strengths at point P due to charges +q and −q respectively.
We know that the direction of electric field due to a point charge is away from positive charge and towards the negative charge. Therefore,
E
1
=
4πϵ
0
1
(r−a)
2
q
(from B to P) and E
2
=
4πϵ
0
1
(r+a)
2
q
(from P to A)
Clearly the directions of electric field strengths
E
1
are
E
2
along the same line but opposite to each other and E
1
>E
2
because positive charge is nearer.
∴ The resultant electric field due to electric dipole has magnitude equal to the difference of E
1
and E
2
direction from B to P i.e.
E=E
1
−E
2
=
4πϵ
0
1
(r−a)
2
q
−
4πϵ
0
1
(r+a)
2
q
=
4πϵ
0
q
[
(r−a)
2
1
−
(r+a)
2
1
]=
4πϵ
0
q
[
(r
2
−a
2
)
2
(r+a)
2
−(r−a)
2
]
=
4πϵ
0
q
(r
2
−a
2
)
2
4ra
=
4πϵ
0
1
(r
2
−a
2
)
2
2(q2a)r
But q.2l=p (electric dipole moment)
∴E=
4πϵ
0
1
(r
2
−a
2
)
2
2pr
....(i)
If the dipole is infinitely small and point P is far away from the dipole, then r>>1, therefore equation (i) may be expressed as
E=
4πϵ
0
1
r
4
2pr
or E=
4πϵ
0
1
r
3
2p
....(ii)
This is the expression for the electric field strength at axial position due to a short electric dipole.
(ii) At a point of equatorial line: Consider a point P on broad side on the position of dipole formed of charges +q and −q at separation 2a. The distance of point P from mid-point (O) of electric dipole is r. Let
E
1
and
E
2
be the electric field strengths due to charges +q and −q of electric dipole.
From fig. AP=BP=
r
2
+a
2
∴
E
=
4πϵ
0
1
r
2
+a
2
q
along B to P
E
2
=
4πϵ
0
1
r
2
+a
2
q
along P to A
Clearly
E
1
and
E
2
are equal in magnitude i.e. ∣
E
1
∣=∣
E
2
∣ or E
1
=E
2
To find the resultant of
E
1
and
E
2
, we resolve them into rectangular components.
Component of
E
1
parallel to AB=E
1
cosθ, in the direction to
BA
Component of
E
1
perpendicular to AB=E
1
sinθ along OP
Component of
E
2
parallel to AB=E
2
cosθ in the direction
BA
∴ Resultant electric field at P is E=E
1
cosθ+E
2
cosθ
But E
1
=E
2
=
4πϵ
0
1
(r
2
+a
2
)
q
From the figure, cosθ=
POB
OB
=
r
2
+a
2
l
=
(r
2
+a
2
)
1/2
l
E=2E
1
cosθ=2×
4πϵ
0
1
(r
2
+a
2
)
q
⋅
(r
2
+a
2
)
1/2
l
=
4πϵ
0
1
(r
2
+a
2
)
3/2
2ql
But q.2l=p= electric dipole moment .... (iii)
∴E=
4πϵ
0
1
(r
2
+a
2
)
3/2
p
If dipole is infinitesimal and point P is far away, we have a<<r, so l
2
may be neglected as compared to r
2
and so equation (iii) gives
E=
4πϵ
0
1
(r
2
)
3/2
P
=
4πϵ
0
1
r
3
P
i.e., electric field strength due to a short dipole at broadside on position
E=
4πϵ
0
1
r
3
p
in the direction parallel to
BA
....(iv)
Its direction is parallel to the axis of dipole from positive to negative charge.
It may be noted clearly from equations (ii) and (iv) that electric field strength due to a short dipole at any point is inversely proportional to the cube of its distance from the dipole and the electric field strength at axial position is twice that at broad-side on position for the same distance.
Important: Note the important point that the electric field due to a dipole at large distances falls off as
r
3
1
and not as
r
2
1
as in the case of a point charge.
