ELectric Dipole

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
  
  E=−2qaPˆ4πϵ0r3
  
                                          (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.
  
  E−q=−qPˆ4πϵ0(r+a)2
  
                                           (4a)
  Pˆ = unit vector along the dipole axis (from -q to +q)
  also
  
  E+q=qPˆ4πϵ0(r−a)2
  
                                           (4b)
• Total field at P is
  E = E_+q + E_-q
  
  
                                           (5)
  for r>>a
  
  E=4qaPˆ4πϵ0r3
  
                                           (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/4πε₀r³
  
  E=−P4πϵ0(r)3
  
  (ii) At point on dipole axis (r>>a)
  
  E=2P4πϵ0(r)3
  
  Note:-
  (i) Dipole field at large distances falls off as 1/r³
  (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 F₁ of magnitude qE, acts on positive charge in direction of electric field and a force F₂ of same magnitude acts on negative charge but it acts in direction opposite to F₁.
  
  
• 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 F₁ about O is
  τ₁= OB × F₁
       = q (OB × E)
  
  Torque of F₂ about O is
  τ₂ = OA × F₂
       = -q(OA × E)
       = q(AO × E)
  
  net torque acting on dipole is
  τ = τ₁ + τ₂
       = 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 q₁ and q₂are located with points having position vectors r₁ and r₂
(1) Find the position vector r₃ where the third charge q₃ should be placed so that force acting on each of the three charges would be equal to zero.
(2) Find the amount of charge q₃

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.