Question

4. What will be the amount of work done to rotate an electric dipole of moment 2.0 x 10-20 C.m through an angle of 180° in an electric field of intensity 1.2 x 104 v.m-l? [4.8 x 10-16]​

Answer 1

Given :-

Dipole Moment = P = 2 × 10-²⁰ C-m

Electric Field = E = 1.2 × 10⁴ V-m

Angle of rotation = Ø = 180°

As we know that,

W = PE - PE CosØ

W = PE (1 - Cos 180°)

W = 2 × 10-²⁰ × 1.2 × 10⁴ [1-(-1)] Cos 180° = -1

W = 4.8 × 10-¹⁶ J

Answer 2

Answer:

Given :-

• An electric dipole of moment 2.0 × 10-²⁰ cm through an angle of 180° in an electric field of intensity 1.2 × 10⁴ v/m.

To Find :-

• What is the amount of work done.

Formula Used :-

$\mapsto \sf\boxed{\bold{\pink{Work\: Done =\: pE(cos\theta_1 - cos\theta_2)}}}$

where,

• p = Dipole Moment
• E = Electric field intensity
• $\sf \theta_1$ = Stable Equilibrium
• $\sf \theta_2$ = Unstable Equilibrium

Solution :-

Given :

$\bigstar\: \: \bf{Dipole\: Moment\: (p) =\: 2.0 \times 10^{- 20}\: cm}$

$\bigstar\: \: \bf{Electric\: Field\: Intensity\: (E) =\: 1.2 \times 10^4\: v/m}$

$\bigstar\: \: \bf{Stable\: Equilibrium\: (\theta_1) =\: 0^{\circ}}$

$\bigstar\: \: \bf{Unstable\: Equilibrium\: (\theta_2) =\: 180^{\circ}}$

According to the question by using the formula we get,

$\leadsto \sf\bf{Work\: Done =\: pE(cos\theta_1 - cos\theta_2)}$

$\longrightarrow \sf Work\: Done =\: (2.0 \times 10^{- 20})(1.2 \times 10^4)\{cos0^{\circ} - cos180^{\circ}\}$

As we know that :

$\bullet\: \: \: \sf\bold{\purple{cos0^{\circ} =\: 1}}$

$\bullet\: \: \sf\bold{\purple{cos180^{\circ} =\: -\: 1}}$

$\longrightarrow \sf Work\: Done =\: 2.0 \times 10^{- 20} \times 1.2 \times 10^4\{1 - (- 1)\}$

$\longrightarrow \sf Work\: Done =\: 2.0 \times 10^{- 20} \times 1.2 \times 10^4(1 + 1)$

$\longrightarrow \sf Work\: Done =\: 2.0 \times 10^{- 20} \times 1.2 \times 10^4(2)$

$\longrightarrow \sf Work\: Done =\: 2 \times 2.0 \times 1.2 \times 10^{- 20} \times 10^4$

$\longrightarrow \sf Work\: Done =\: 4 \times 1.2 \times 10^{- 20} \times 10^4$

$\longrightarrow \sf Work\: Done =\: 4.8 \times 10^{- 20} \times 10^4$

$\longrightarrow \sf Work\: Done =\: 4.8 \times 10^{(- 20 + 4)}$

$\longrightarrow \sf\bold{\red{Work\: Done =\: 4.8 \times 10^{- 16}\: J}}$

${\small{\bold{\underline{\therefore\: The\: amount\: of\: work\: done\: is\: 4.8 \times 10^{- 16}\: J\: .}}}}$