Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

Derive the expression for electric field on the axis of an electric dipole.

Accepted Answer

We have been asked to derive an expression for electric field intensity due to an electric dipole at a point on its axial line.

Now, let AB be an electric dipole of two-point charges $- q$ and $+ q$ separated by a small distance $2 d$ .

Also, consider that P is a point along the axial line of the dipole at a distance $r$ from the midpoint O of the electric dipole.

Now, for better understanding refer the figure below:

The electric field at any point due to a charge $q$ at a distance $x$ is given by-

$E = \frac{1}{4 π \in_{\circ}} . \frac{q}{x^{2}}$

Now, the electric field at the point P due to $+ q$ charge placed at B is given by-

$E_{1} = \frac{1}{4 π \in_{\circ}} . \frac{q}{{(r - d)}^{2}}$ (along BP)

Here, $(r - d)$ is the distance of point P from charge $+ q$ .

Also, the electric field at point due to $- q$ charge placed at A.

$E_{2} = \frac{1}{4 π \in_{\circ}} . \frac{q}{{(r + d)}^{2}}$ (along PA)

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

The resultant electric field at P is-

$E = E_{1} + (- E_{2})$

Putting the values; we get-

$E = [\frac{1}{4 π \in_{\circ}} . \frac{q}{{(r - d)}^{2}} - \frac{1}{4 π \in_{\circ}} . \frac{q}{{(r + d)}^{2}}]$ (along BP)

Simplifying further we get-

$E = \frac{q}{4 π \in_{\circ}} [\frac{1}{{(r - d)}^{2}} - \frac{1}{{(r + d)}^{2}}] E = \frac{q}{4 π \in_{\circ}} [\frac{{(r + d)}^{2} - {(r - d)}^{2}}{{(r - d)}^{2} {(r + d)}^{2}}]$

Now, we can write ${(r - d)}^{2} {(r + d)}^{2} = {(r^{2} - d^{2})}^{2}$

So, we get,

$E = \frac{q}{4 π \in_{\circ}} [\frac{{(r)}^{2} + {(d)}^{2} + 2 r d - {(r)}^{2} - {(d)}^{2} + 2 r d}{{(r - d)}^{2} {(r + d)}^{2}}] E = \frac{q}{4 π \in_{\circ}} [\frac{4 r d}{{(r^{2} - d^{2})}^{2}}]$

Now, if the point P is far away from the dipole, then $d ≪ r$

So, the electric field will be-

$E = \frac{q}{4 π \in_{\circ}} [\frac{4 r d}{{(r^{2})}^{2}}] = \frac{q}{4 π \in_{\circ}} . \frac{4 r d}{r^{4}} = \frac{q}{4 π \in_{\circ}} . \frac{4 d}{r^{3}}$ (along BP)

Also, electric dipole moment $p = q \times 2 d$ , so the expression will now become-

$E = \frac{q}{4 π \in_{\circ}} . \frac{2 (2 d)}{r^{3}} = \frac{1}{4 π \in_{\circ}} . \frac{2 p}{r^{3}}$

E acts in the direction of the dipole moment.

Therefore, the expression for the electric field intensity due to an electric dipole at a point on its axial line is

$E = \frac{q}{4 π \in_{\circ}} . \frac{2 (2 d)}{r^{3}} = \frac{1}{4 π \in_{\circ}} . \frac{2 p}{r^{3}}$

Derive the expression for electric field on the axis of an electric dipole.

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