Question

a) Define an ideal electric dipole. Give an example.

b) Derive an expression for the torque experienced by an electric

dipole in a uniform electric field. What is net force acting on this

dipole.

c) An electric dipole of length 2cm is placed with its axis making an

angle of 600 with respect to uniform electric field of 105N/C.

If it experiences a torque of 8√3 Nm, calculate the (i) magnitude of

charge on the dipole, and its potential energy.​

Answer 1

Explanation:

Define an ideal electric dipole. Give an example.

We can think of an ideal dipole in which size 2α→0 and charge q→∞ in such a way that the dipole moment, p=q×2α a has a finite value, such a dipole of negligibly small size is called an ideal or point dipole.

Derive an expression for the torque experienced by an electric

Torque on Dipole placed in uniform electric field ! |p| = 2aq is placed in a uniform electrinc field E at an angle ϴ. A force F₁ = qE will act on positive charge and F₂ = -qE on negative charge . Since, F₁ and F₂ are equal in magnitude but opposite in direction.

dipole in a uniform electric field. What is net force acting on this dipole.

Force acting on the dipole

Thus, net force acting on a dipole in a uniform electric field is zero.

An electric dipole of length 2cm is placed with its axis making an

An electric dipole of length 2 cm is placed with its axis making an angle of `60^(@)` to a uniform electric field of `10^(5)NC^(-1)` if its experiences a torque is `8sqrt(3)`Nm, calculate the (i). Magnitude of the charge on the dipole and (ii). potential energy of the dipole.

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Answer 2

Answer:

Concept:

Ideal Electric dipole, torque experienced by an electric dipole.

Explanation:

a) Definition:

We get an ideal dipole or point dipole when the charge q increases and the distance 2a decreases, but the product  $|\vec{P}|=$ = q.(2a) remains constant.

Some protein biomolecules behave like a perfect electric dipole.

$where,\\ \mathbf{q}= magnitude of either charge \\\mathbf{2} \mathbf{a}=distance between charges \\Dipole moment (|\vec{P}|)=\mathrm{q} \cdot 2 \mathrm{a}$

b) Expression for the torque experienced by an electric dipole.

As shown in the attached image,

Since E is uniform, the $(qE-qE) = 0 Due \ to \ couple, dipole \ is \ rotated \ in \ anti-clockwise \ direction.\\\\Draw AC \perp \vec{E} \\ \therefore \perp_{r} distance between the forces = arm of couple =\mathrm{AC} \\\\As, \tau= moment of the couple= force x arm of couple\\ =f \times A c \\ =f \times A B \sin \theta \\ =f \times 2 a \sin \theta \\ =(q E \times 2 a) \sin \theta \\ =(q \times 2 a) E \sin \theta \\or, \tau=P E \sin \theta \\In vector form, \vec{\tau}=\vec{p} \times \vec{E}$

c)(i)

$As \ Given:- \ \\\\ \vec{E}=10^{5} \mathrm{~N} / \mathrm{C} \begin{aligned}&\text { Za }=2 \mathrm{~cm} \\&\theta=60^{\circ} \\&\tau=8 \sqrt{3} \mathrm{Nm} \\&\because \tau=P E \sin \theta \\&8 \sqrt{3}=2 a \times q \times 10^{5} \sin 60^{\circ} \\&8 \sqrt{3}=2 \times 10^{-2} \times q \times 10^{5} \times \frac{\sqrt{3}}{2} \\&\therefore q=8 \times 10^{-3} \mathrm{C}\end{aligned}$

(ii)

$Potential energy, P.E =-P E \cos \theta \begin{aligned}&=-2 a \times q \times E \times \cos 60^{\circ} \\&=-8 \times 10^{-3} \times 2 \times 10^{-2} \times 10^{5} \times \frac{1}{2} \\&\therefore \text { P.E. }=\mathbf{- 8 J}\end{aligned}$

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