9. Electric Dipole
- Electric dipole is a pair of equal and opposite charges, +q and -q, separated by some distance 2a.
- Total charge of the dipole is zero but electric field of the dipole is not zero as charges q and -q are separated by some distance and electric field due to them when added is not zero.
(A)Field of an electric dipole at points in equitorial plane
- We now find the magnitude and direction of electric field due to dipole.
- P point in the equitorial plane of the dipole at a distance r from the centre of the dipole. Then electric field due to +q and -q are
$$\boldsymbol{E_{-q}}=\frac{-q\boldsymbol{\widehat{P}}}{4\pi \epsilon _{0}(r^{2}+a^{2})}$$
(1a)
$$\boldsymbol{E_{+q}}=\frac{q\boldsymbol{\widehat{P}}}{4\pi \epsilon _{0}(r^{2}+a^{2})}$$
(1b)
and they are equal
Pˆ = unit vector along the dipole axis (from -q to +q)
- From fig we can see the direction of E_{+q} and E_{-q}. Their components normal to dipole cancel away and components along the dipole add up.
- Dipole moment vector points from negative charge to positive charge so in vector form.
E = -(E_{+q} + E_{-q}) cos θ
(2)
At large distances (r>>a), above equation becomes
$$\boldsymbol{E}=\frac{-2qa\boldsymbol{\widehat{P}}}{4\pi \epsilon _{0}r^{3}} $$
(3)
and they are equal
(B) Field of an electric dipole for points on the axis
- Let P be the point at a distance r from the centre of the dipole on side of charge q. as shown in the fig.
$$\boldsymbol{E_{-q}}=\frac{-q\boldsymbol{\widehat{P}}}{4\pi \epsilon _{0}(r+a)^{2}}$$
(4a)
Pˆ = unit vector along the dipole axis (from -q to +q)
also
$$\boldsymbol{E_{+q}}=\frac{q\boldsymbol{\widehat{P}}}{4\pi \epsilon _{0}(r-a)^{2}}$$
(4b)
- Total field at P is
E = E_{+q} + E_{-q}
(5)
for r>>a
$$\boldsymbol{E}=\frac{4qa\boldsymbol{\widehat{P}}}{4\pi \epsilon _{0}r^{3}} $$
(6)
- For equation (3) and (6) charge q and dipole separation 2a appear in combination qa. This leads us to define dipole moment vector P of electric dipole. Thus, electric dipole moment
P = q × 2a Pˆ (7)
- Unit of dipole moment is Coulomb's meter (Cm).
- In terms of electric dipole moment, field of a dipole at large distances becomes
(i) At point on equitorial plane (r>>a)
E = -P/4πε_{0}r^{3}
$$\boldsymbol{E}=\frac{-\boldsymbol{P}}{4\pi \epsilon _{0}(r)^{3}}
$$
(ii) At point on dipole axis (r>>a)
$$\boldsymbol{E}=\frac{2\boldsymbol{P}}{4\pi \epsilon _{0}(r)^{3}}
$$
Note:-
(i) Dipole field at large distances falls off as 1/r^{3}
(ii) Both the direction and magnitude of dipole an angle between dipole moment vector P and position vector r
(C) Dipole in a uniform external field
- Consider a dipole in a uniform electric field E whose direction makes an angle θ with dipole axis (line joining two charges)
- Force F_{1} of magnitude qE, acts on positive charge in direction of electric field and a force F_{2} of same magnitude acts on negative charge but it acts in direction opposite to F_{1}.
- Resultant force on dipole is zero, but since two forces do not have same line of action they constitute a couple.
- We now calculate torque (r × F) of these forces about zero.
Torque of F_{1} about O is
τ_{1}= OB × F_{1}
= q (OB × E)
Torque of F_{2} about O is
τ_{2} = OA × F_{2}
= -q(OA × E)
= q(AO × E)
net torque acting on dipole is
τ = τ_{1} + τ_{2}
= q(OB+ AO) × E
= q (AB × E)
AB = 2a and p = 2qa (dipole moment)
τ =p × E
- Direction of torque is perpendicular to the plane containing dipole axis and electric field.
- Effect of torque is to rotate the dipole to a position in which dipole moment p is parallel to E the electric field vector is shown above in figure b and for uniform electric field dipole is in equilibrium in this position.
- magnitude of this torque is
τ = |τ| = pE sinθ
Examples
Question 1
Two point charges q_{1} and q_{2}are located with points having position vectors r_{1} and r_{2}
(1) Find the position vector r_{3} where the third charge q_{3} should be placed so that force acting on each of the three charges would be equal to zero.
(2) Find the amount of charge q_{3}
Question 2
Consider a thin wire ring of radius R and carrying uniform charge density λ per unit length.
(1) Find the magnitude of electric field strength on the axis of the ring as a function of distance x from its centre.
(2) What would be the form of electric field function for x>>R.
(3) Find the magnitude of maximum strength of electric field.
Question 3
Two equally charged metal balls each of mass m Kg are suspended from the same point by two insulated threads of length l m long. At equilibrium, as a result of mutual separation between balls, balls are separated by x m. Determine the charge on each ball.
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Class 12 Maths
Class 12 Physics