Name: Electric Field of a Charged Plane Sheet
Uploaded: 2019-07-19
Duration: 6 min 58 s
Description: Another distribution of charge. This time on an infinite plane sheet. Let’s see once again how Gauss’s law comes to the rescue in electric field calculations.

Question

Define electric dipole moment. Derive an expression for the electric field at a point on equatorial (broad-side-on) position of an electric dipole.

Accepted Answer

Electric dipole moment:

The product of either change $(q)$ of electric dipole and the distance between them $(l)$ is referred to be Electric dipole moment. That is, $\vec{p} = q \times l$ .

Electric field intensity on equatorial line of dipole:

Considering an electric dipole with a pair of charge $+ q$ and $- q$ separated by a small distance $2 a$ .

Considering an equatorial line of dipole at a point $P$ . At this point the net electric intensity will be due to charge $+ q$ and $- q$ .

Assuming $\vec{E_{1}}$ be the electric intensity at point $P$ due to $+ q$ charge of the electric dipole.

Then, $\vec{E_{1}} = \frac{1}{4 π ε_{\circ}} \frac{q}{{(A P)}^{2}}$ along $P C$ .

Here, ${(A P)}^{2} = r^{2} + a^{2}$ by substituting this in above equation.

$\vec{E_{1}} = \frac{1}{4 π ε_{\circ}} \frac{q}{r^{2} + a^{2}}$ ................................... $(1)$

Here, $∠ P A B = ∠ P B A = θ$

Then the rectangular components of $\vec{E_{1}}$ are:

Vertical component: $\vec{E_{1}} \sin θ$ along $P L$

Horizontal component: $\vec{E_{1}} \sin θ$ along $P R$

Assuming $\vec{E_{2}}$ be the electric intensity at point $P$ due to $- q$ charge of the electric dipole.

Then, $\vec{E_{2}} = \frac{1}{4 π ε_{\circ}} \frac{q}{{(P B)}^{2}}$ along $P D$ .

Here, ${(P B)}^{2} = r^{2} + a^{2}$ by substituting this in above equation.

$\vec{E_{1}} = \frac{1}{4 π ε_{\circ}} \frac{q}{r^{2} + a^{2}}$ ................................... $(2)$

Then the rectangular components of $\vec{E_{2}}$ are:

Vertical component: $\vec{E_{2}} \sin θ$ along $P M$

Horizontal component: $\vec{E_{2}} \cos θ$ along $P R$

The net electric intensity at point $P$ : $\vec{E_{1}} \sin θ + \vec{E_{1}} \cos θ + \vec{E_{2}} \sin θ + \vec{E_{2}} \cos θ$

Here, the vertical component of $\vec{E_{1}}$ and $\vec{E_{2}}$ that is $\vec{E_{1}} \sin θ$ and $\vec{E_{2}} \sin θ$ are of same magnitude but in opposite direction so they will cancel out each other.

So the net electric intensity at point $P$ will be $\vec{E_{n e t}} = \vec{E_{1}} \cos θ + \vec{E_{2}} \cos θ = 2 \vec{E_{1}} \cos θ$

$\Rightarrow \vec{E_{n e t}} = 2 \frac{1}{4 π ε_{\circ}} \frac{q}{r^{2} + a^{2}} \cos θ$

Substituting, $\cos θ = \frac{a}{\sqrt{a^{2} + r^{2}}}$ in the above equation.

$\vec{E_{n e t}} = 2 \frac{1}{4 π ε_{\circ}} \frac{q}{r^{2} + a^{2}} \frac{a}{\sqrt{r^{2} + a^{2}}} = \frac{1}{4 π ε_{\circ}} \frac{q (2 a)}{{(r^{2} + a^{2})}^{\frac{3}{2}}} = \frac{1}{4 π ε_{\circ}} \frac{p}{{(r^{2} + a^{2})}^{\frac{3}{2}}}$

Here, $p$ is the electric dipole moment where $p = q (2 a)$ .

$\vec{E_{n e t}}$ is along $P R$ .

If $r > > a$ then $\vec{E_{n e t}} = \frac{1}{4 π ε_{\circ}} \frac{p}{r^{3}}$

Hence, the expression for the electric field at a point on equatorial position of an electric dipole has been given.

Embibe Experts Solutions for Chapter: Electric Charges and Fields, Exercise 1: Jharkhand Board-2019

Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Electric Charges and Fields, Exercise 1: Jharkhand Board-2019

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