- 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**) cos θ_{-q}

(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**of magnitude q_{1}**E**, acts on positive charge in direction of electric field and a force**F**of same magnitude acts on negative charge but it acts in direction opposite to F_{2}_{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**= 2q**a**(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

(1) Find the position vector

(2) Find the amount of charge q

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.

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.

- Electric Charge , Basic properties of electric charge and Frictional Electricity
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- Electrical and electrostatic force
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- Coulomb's Law (with vector form)
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- Principle Of Superposition
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- Electric Field
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- Calculation of Electric Field
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- Electric Field Lines
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- Electric Flux
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- Electric Dipole

Class 12 Maths Class 12 Physics

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