Question

derive the expansion for the electric field at point on the axial line and equatorial line of electric dipole?

Answer 1

Condition:

Derive the expansion for the electric field at point on the axial line and equatorial line of electric dipole?

Solution and answer:

Electrostatic dipole is a system of two adjacent equal magnitude charges $+q$ and $-q$ (pic 1). Dipoles are characterized by a dipole moment

$$p = q l = l _ {0} q l,$$

where $l$ -vector pointing from the negative to the positive charge, the absolute value is equal to the distance between the charges $l$ , and $l_0$ is the unit vector corresponding to the vector $l(l = l_o l)$ .

If you bring together the charges, while increasing their value so that the vector $p$ remained unchanged, we obtain a point or ideal dipole with the same momentum.

We calculate the electrostatic field of the dipole. We introduce a spherical coordinate system $r, \theta, \varphi$ , so that the polar axis passes through both charges and the origin is equidistant from them (pic 1). The potential created by the dipole, we find due to the principle of superposition as the sum of the potentials produced by the charges $+q$ and $-q$ :

pic 2

$$u = \frac {q}{4 \pi \varepsilon} \left(\frac {1}{R _ {1}} - \frac {1}{R _ {2}}\right),$$

where $R_{1}$ and $R_{2}$ - the distances from the charges $+q$ and $-q$ to the point at which the potential is calculated (pic 2):

$$R _ {1} = \sqrt {r ^ {2} + \left(\frac {l}{2}\right) ^ {2} - l r \cos \theta}, \quad R _ {2} = \sqrt {r ^ {2} + \left(\frac {l}{2}\right) ^ {2} + l r \cos \theta}.$$

In the calculation of the field, we assume that the distance $r$ from the center of the dipole to the observation point is large compared to the distance between the charges $l$ . Under this condition, we have the following approximate equality

$$r \mp \frac {l \cos \theta}{2} \cong \left\{ \begin{array}{l} R _ {1} \\ R _ {2} \end{array} , \frac {1}{R _ {1}} - \frac {1}{R _ {2}} \cong \frac {l \cos \theta}{2}. \right.$$

In this case (2) takes the form

$$u = \frac {q l \cos \theta}{4 \pi \varepsilon r ^ {2}} = \frac {(p , r _ {0})}{4 \pi \varepsilon r ^ {2}}$$

where $r_o$ - coordinate unit vector of variable $r$ . To determine the electric field we use $E = -\nabla u$ . The expression for the gradient in spherical coordinates is

$$\nabla \mathrm {f} = \frac {\partial \mathrm {f}}{\partial \tau} r _ {0} + \frac {1}{r} \frac {\partial \mathrm {f}}{\partial \theta} \theta_ {0} + \frac {1}{\mathrm {r s i n} \theta} \frac {\partial \mathrm {f}}{\partial \varphi} \varphi_ {0}. \mathrm {U s i n g} \frac {d u}{d \varphi} = 0, \mathrm {w e o b t a i n}$$ $$E = \frac {q l}{4 \pi \varepsilon r ^ {2}} (r _ {0} 2 \cos \theta + \theta_ {0} \sin \theta).$$

Directions of the unit vectors $r_0, \theta_0$ and $\varphi_0$ are shown in pic 1. As can be seen, the vector of the electric field created by an electrostatic dipole, is independent of the angle $\varphi$ (field has axial symmetry) and has two components:

pic 3

$$E _ {r} = \frac {q l \cos \theta}{2 \pi \varepsilon r ^ {3}}, \quad E _ {0} = \frac {q l \sin \theta}{4 \pi \varepsilon r ^ {3}}.$$

The field lines are shown in pic 3.

For the electric field at the point on the axial line $(\theta = 0)$ :

$$E _ {r} = \frac {q l}{2 \pi \varepsilon r ^ {3}}, \quad E _ {0} = 0.$$

For the electric field at the point on equatorial line $(\theta = \frac{\pi}{2})$ :

$$E _ {r} = 0, \quad E _ {0} = \frac {q l}{4 \pi \varepsilon r ^ {3}}.$$