Electric Dipole - Electric Field on Equatorial (Transversal) Line

Description:

To calculate electric field created by a dipole on the equatorial or transversal line (on the line perpendicular to the line joining the two charges),

• All the measurement of distances are to be taken from the centre(O). Which is the mid point of the dipole

• Let the distance between O to +q and  O to –q be ‘l’. So, total length between +q and –q will be ‘2l’.

• Take a point ‘p’ on the equatorial  line at the distance ‘r’ from the centre of the dipole as shown in figure.

• Let the angle made by both the charges with respect to the point P be ‘θ’.

Direction of electric field on the equatorial line of the dipole

• By coulomb’s law The direction of the electric field at P due to charge +q will be away from itself whereas due to point charge -q , it will be towards itself (-q), in the line joining the two charges as shown in figure.

Note −

Since  the magnitude of electric field E_+q and E_-q  are equal so the length of electric field vector are kept same as shown in figure.

To calculate electric field at point ‘P’ due to the two point charges of dipole.

By using the formula for electric field due to point charge,

Electric field due to +q E_+q = +1/4πε₀ q/(r² + l²) .....(1)

Since, by Pythagoras theorem the distance between P and +q is (r² + l²)^1/2

Electric field due to -q E_-q = -1/4πε₀ q/(r² + l²)

(Electric field due to +q will be positive and electric field due to -q will be negative)

Since  the magnitude of electric field E_+q and E_-q are equal so E_-q = E_+q = E.

To find resultant electric field,

Resolving the electric field due to both the charges into two components.

From figure −

• Sinθ component of electric field due to both the charges cancelled out. As they are equal and opposite in direction.

• Cosθ component of electric field due to both the charges added up. As they are in same  direction.

So, the resultant electric field (E_R) due to both the charges will be −

E_R = E Cosθ + E Cosθ

E_R = 2 E Cosθ ......(2)

Putting (1) in (2)

E_R =  1/4πε₀  2q/(r² + l²) 1/(r² + l²)^1/2 [from figure Cosθ = 1/(r² + l²)^1/2]

We know that the dipole moment of a dipole is P = 2ql, so on putting it in above equation we get,

E_R = 1/4πε₀ P/(r² + l²)^3/2

Since, the dipole is very small so ‘l’ is also very small as compared to the distance ‘r’.

So, on neglecting ‘l’ with respect to ‘l’ we get −

E_R = 1/4πε₀ P/r³

On comparing the electric field due to a dipole in axial position and in equatorial position we find that −

Electric field at axial position (E_axial) is double the Electric field at equatorial position (E_equitorial) when ‘r’ is equal.

(E_axial) = 2(E_equatorial) (Only when ‘r’ is equal)

Resultant direction of electric field in transversal line

The direction of electric field on the equatorial line is always opposite to the direction of dipole moment.

On comparing the electric field due to a point charge and electric field due to a dipole in equatorial position we find that −

• As ‘r’ increases  the Electric field decreases more  in case of dipole as compared to electric field due to a point charge. In dipole E ∝ 1/r³ For point charge E ∝ 1/r²

So we can say − Dipole Electric field depletes faster then electric field by a point charge.