Question

Show that intensity of electric field at a point in broadside position of an electric dipole is given by:

E = `(1/(4 pi epsilon_0)) p/((r^2 + l^2)^(3//2))`

Where the terms have their usual meaning.

Show that intensity of electric field E at a point in broadside on position is given by:

E = `(1/(4 pi epsilon_0)) p/((r^2 + l^2)^(3//2))`,

where the terms have their usual meaning.

Answer 1

Consider a dipole of length 2l and moment `vec p`.

From the figure, resultant electric field intensity at C:

`vec E = vec E_A + vec E_B`

`|vec E_A| = 1/(4 pi epsilon_0) q/(r^2 + l^2)`

`|vec E_B| = 1/(4 pi epsilon_0) q/(r^2 + l^2)`

Sine components of `vec E_A` and `vec E_B` get cancelled each other as `|vec E_A| = |vec E_B|`.

The cosine components get added up to give the resultant field.

i.e., E = E_A cos θ + E_B cos θ

= `1/(4 pi epsilon_0) * q/(r^2 + l^2) cos theta + 1/(4 pi epsilon_0) * q/(r^2 + l^2) cos theta`

= `2 * 1/(4 pi epsilon_0) * q/(r^2 + l^2) cos theta`

= `2 xx 1/(4 pi epsilon_0) * q/(r^2 + l^2) xx l/(r^2 + l^2)^(1//2)`

= `1/(4 pi epsilon_0) (q(2 l))/((r^2 + l^2)^(3//2))`

E = `1/(4 pi epsilon_0) p/((r^2 + l^2)^(3//2))`    ...[as p = q(2l)]

The above expression gives the magnitude of the field. The direction of the electric field E at C is opposite to the direction of the dipole moment `vec p`.