## Potential And Field

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# Potential and Field due to an Electric Dipole

A dipole consisting of two charges + q and – q separated by a distance 2d.Dipole moment of the electric dipole = p = q x 2d.

Direction of p is form – q to + q.

P is a point at a distance r from the centre O.

∟POB = θ. Let us calculate the potential at P.

AC and BD are drawn perpendicular to OP.

Potential at P due} = __

___1_____ (q/PB)

to the charge + q 4πε

_{0}

Potential at P due} = __

___1_____ (q/PA)

to the charge - q 4πε

_{0}

Let us assume r >> d. Then

PA = PC = PO + OC = r + d cos θ

PB = PD = PO + OD = r – d cos θ

The resultant potential at P is

V = __

__1____ [_____

__q________ - _____

__q_______]

4πε

_{0}(r – d cos2θ) ( r + d cos θ)

= _

___1____ [_

__2d cos θ _______]

4πε

_{0}(r

^{2}– d

^{2}cos

^{2}θ)

d

^{2}cos

^{2}θ << r

^{2}and so it can be neglected. Also, q X 2d = P.

. : V = __

__1____

__p cos θ__)

_______4πε

_{0}r

^{2 }

or V = __

__1____

__p.r__

_______4πε

_{0}r

^{2}

Here, r is the unit vector along r. θ is the angle between p and r (i.e., OP).

Special cases (i) When the point P lies on the axial line of the dipole on the side of the positive charge, θ = 0 and cos θ = 1.

V = __

___1_____

__p__

______4πε

_{0}r

^{2 }

(ii) When the point p lies on the axial line of the dipole on the side of the negative charge, θ = 180 and cos θ = - 1.

V = ___

__1_____

__(-p)__

______4πε

_{0}r

^{2}

(iii) When the point P lies on the equatorial line of the dipole, θ = 90 and cos θ = 0. Therefore V = 0.

**Electric Field.**The field at P is calculated using the relation E = - Δ V.

The components of the field along the radius vector r is

Er = - dV =

__- d__(__

___1_____

__p cos θ__) = __

___1_____

__2p cos θ__

dr dr 4πε

_{0 }r

^{2}_______4πε

_{0}r

^{3 }

Along a direction perpendicular to r, field component

E

_{θ}= -

__1 dV__=

__- 1__[d (__

___1_____

__p cos θ__) ] =

_____1_____

__p sin θ__

r d θ r dθ 4πε

_{0}r

^{2}_______ 4πε

_{0}r

^{3}

The magnitude of the resultant field at P is

E = √(E

^{2}

_{r}+ E

^{2}

_{θ}) = ___

__1_____

__p__(√(3cos

^{2}θ + 1)

___________ 4πε0 r3

The angle Ф which the resultant field E makes with r is

Ф = tan

^{-1}(E

_{θ}/ E

_{r}) = tan

^{-1}(1/2 tan θ)

**Special cases.**(i) If P is on the axial, line, θ = O.

**E**is in the direction of p and has magnitude 2p / (4πε

_{0}r

^{3})

(ii) If P is on the equatorial line, θ = 90.

**E**is in the direction of – p and has magnitude p / (4πε

_{0}r

^{3}).

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