Name: Equipotential Surfaces and Its Properties
Uploaded: 2019-07-19
Duration: 7 min 22 s
Description: The surface over which the electric potential is the same everywhere is called an equipotential surface. Let's more about it and its properties in this video.

Question

Derive an expression for electric field intensity due to an electric dipole at a point on its axial line.

Accepted Answer

Electric field due to an electric dipole at a point on its axial line: AB is an electric dipole of two point charges −q and +q separated by small distance 2d. P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.

The electric field at the point P due to +q placed at B is,

$E_{1} = \frac{1}{4 π ε_{0}} \frac{q}{{(r - d)}^{2}} (a l o n g B P)$

The electric field at the point P due to −q placed at A is,

$E_{2} = \frac{1}{4 π ε_{0}} \frac{q}{{(r + d)}^{2}} (a l o n g P A)$

Therefore, the magnitude of resultant electric field (E) acts in the direction of the vector with a greater, magnitude. The resultant electric field at P is

$E = E_{1} + (- E_{2} ) \Rightarrow E = [ \frac{1}{4 π ε_{0}} \frac{q }{ {(r - d)}^{2}} - \frac{1}{4 π ε_{0}} \frac{q }{ {(r + d)}^{2}} ] a l o n g B P \Rightarrow E = \frac{q}{4 π ε_{0}} [\frac{1}{{(r - d)}^{2}} - \frac{1}{{(r + d)}^{2}} ] a l o n g B P \Rightarrow E = \frac{q}{4 π ε_{0}} [\frac{4 r d}{{(r^{2} - d^{2})}^{2}} ] a l o n g B P$

If the point P is far away from the dipole, then d≪r

$∴ E = \frac{q}{4 π ε_{0}} - \frac{4 r d}{r^{4}} = \frac{q}{4 π ε_{0}} \frac{4 d}{ r^{3}} \Rightarrow E = \frac{1}{4 π ε_{0}} \frac{2 p}{r^{3}} a l o n g B P$

[∵ Electric dipole moment p=q×2d]

E acts in the direction of dipole moment.

Derive an expression for electric field intensity due to an electric dipole at a point on its axial line.

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