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 an expression for the magnitude of electric field intensity at any point along the equatorial line of a short electric dipole. Give the direction of intensity at that point. For a short dipole what is the ratio of magnitude of electric field intensities at two equidistant points from the centre of dipole, one along the axial line and another on the equatorial line ?

Accepted Answer

Consider an electric dipole AB consisting charges $+ q$ and $- q$ separated by distance $2 l$ and pole strength $q$ . Let, P be the point on the equatorial line at distance $d$ from mid-point of the dipole i.e. O as shown in figure 1.

Now, the electric field, $E_{1}$ at point P due to the North Pole along AP is,

$E_{1} = \frac{1}{4 π \in_{\circ}} \frac{q}{A P^{2}} E_{1} = \frac{1}{4 π \in_{\circ}} \frac{q}{{(d^{2} + l^{2})}^{2}} . . . . . . . . . (1) (∵ A P^{2} = O A^{2} + O P^{2})$

Similarly, the electric field, $E_{2}$ at point P due to the South Pole along PB is,

$E_{2} = \frac{1}{4 π \in_{\circ}} \frac{q}{P B^{2}} E_{2} = \frac{1}{4 π \in_{\circ}} \frac{q}{{(d^{2} + l^{2})}^{2}} . . . . . . . . . (2) (∵ P B^{2} = O B^{2} + O P^{2})$

From equations (1) and (2), we get

$E_{1} = E_{2} = E (s a y)$

Now, the net electric field $E_{P}$ at P due to the dipole is

$E_{P} = E_{1} + E_{2} = 2 E . . . . . . . . . . . . . . . . . . . . . . . . (3)$

Resolving $E_{1}$ and $E_{2}$ into horizontal and vertical components, as shown in figure 2. The vertical components will be cancelled and horizontal will be added together. So, we can write

$E_{P} = 2 E \cos θ$ (from eq.(3))

$E_{p} = 2 (\frac{1}{4 π \in_{\circ}} \frac{q}{{(d^{2} + l^{2})}^{2}} . \frac{1}{\sqrt{(d^{2} + l^{2})}}) (∵ \cos θ = \frac{1}{\sqrt{(d^{2} + l^{2})}})$

For a short dipole, $d < < l$ , therefore

$E_{P} = \frac{1}{4 π \in_{\circ}} \frac{2 q l}{l^{3}} E_{P} = \frac{1}{4 π \in_{\circ}} \frac{p}{l^{3}} (∵ p = 2 q l)$

For a point on the equatorial line of dipole,
The direction of electric, field $\vec{E}$ is along a line parallel to the axis of dipole directed opposite to the direction of dipole moment $\vec{P}$ .

Electric field due to an electric dipole at points situated at a distance $r$ along its axial line is given as,

$E_{a x i a l} = \frac{2 p}{4 π \in_{\circ} r^{3}} . . . . . . . . . . . . . . (4)$

Electric field due to an electric dipole at points situated at a distance $r$ along its equatorial line is given as,

$E_{e q u .} = \frac{p}{4 π \in_{\circ} r^{3}} . . . . . . . . . . . . . . (5)$

From (4) & (5)

$\frac{E_{a x i a l}}{E_{e q u .}} = \frac{2}{1}$

Therefore, ratio is $2 : 1$

Important Questions on Electric Charges and Fields

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