#### Question

(a) Derive an expression for the electric field E due to a dipole of length '2a' at a point distant r from the centre of the dipole on the axial line.

(b) Draw a graph of E versus r for r >> a.

(c) If this dipole were kept in a uniform external electric field E_{0}, diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases.

#### Solution

**(a) Electric Field on Axial Line of an Electric Dipole: **

Let P be at distance *r* from the centre of the dipole on the side of charge *q*. Then,

`E_(-q)=q/(4piepsilon_0(r+a)^2)hatp`

Where, `hatp` is the unit vector along the dipole axis (from − *q* to *q*). Also,

`E_(+q)=q/(4piepsilon_0(r-a)^2)hatp`

The total field at P is

`E=E_(+q)+E_(-q)=q/(4piepsilon_0)[1/(r-a)^2-1/(r+a)^2]hatp`

`E=q/(4piepsilon_0)(4ar)/(r^2-a^2)^2hatp`

for r >> a

`E=(4qa)/(4piepsilon_0r^3)hatp` (r >> a)

`E=(2p)/(4piepsilon_0r^3)` `[because vecp=qxxvec(2a)hatp]`

**(b)** For r >> a, `E prop 1/r^3`

**(c) Position of dipole in stable equilibrium:** In stable equilibrium, dipole aligns itself in the direction of external electric field.

Therefore, angle (θ) between `vecP` and `vecE_0` is 0 and we know torque acting on the dipole in an external field is given by the expression

τ = PEsinθ

As, θ is 0° for stable equilibrium

Therefore, τ = PEsinθ

τ= PEsin0°

τ = 0

Position of dipole in unstable equilibrium: In unstable equilibrium, dipole aligns itself in the direction opposite to the direction of external electric field.

τ = PEsinθ

As, θ is 180° for unstable equilibrium

Therefore, τ = PEsinθ

τ= PEsin180°

τ = 0