📘 WORKSHEET: V due to an Electric Dipole, Special cases, Rotational work done by a Dipole in an External Electric Field, Rotational Potential Energy stored in a Dipole in an External Electric Field

🔹 SECTION A: THEORY QUESTIONS (1–10)

Q1. Define electric dipole moment. Write its SI unit and direction.

Q2. Write the expression for electric potential due to a dipole at a general point.

Q3. What is the electric potential at a point on the axial line of a dipole?

Q4. What is the electric potential at a point on the equatorial line of a dipole?

Q5. State whether potential is zero or non-zero on the equatorial line and explain why.

Q6. How does electric potential due to a dipole vary with distance?

Q7. Write the expression for work done in rotating a dipole in a uniform electric field.

Q8. Define potential energy of a dipole in an external electric field.

Q9. Write expressions for:

Stable equilibrium
Unstable equilibrium

Q10. Draw qualitative variation of potential energy vs angle between dipole and field.

🔹 SECTION B: NUMERICAL QUESTIONS (11–20)

Q11. A dipole has charges ±2 μC separated by 4 cm. Find its dipole moment.

Q12. Find the potential at a point 20 cm away on the axial line of a dipole of moment
p = 4 × 10⁻⁸ C·m.

Q13. Find the potential at a point on the equatorial line at distance 10 cm from the center of a dipole.

Q14. A dipole is placed in a uniform electric field of 5 × 10⁴ N/C. If p = 2 × 10⁻⁶ C·m, find the torque when θ = 30°.

Q15. Calculate the work done in rotating a dipole from 0° to 90° in a uniform field.
Given: p = 3 × 10⁻⁶ C·m, E = 10⁵ N/C.

Q16. Find the potential energy of a dipole at 60° with the field.
Given: p = 5 × 10⁻⁶ C·m, E = 2 × 10⁴ N/C.

Q17. What is the change in potential energy when dipole rotates from 0° to 180°?

Q18. A dipole of moment 10⁻⁸ C·m is placed in a field 10⁵ N/C. Find maximum and minimum potential energy.

Q19. At what angle is potential energy zero?

Q20. If work done in rotating dipole from 0° to θ is W, derive relation between W and θ.

✅ ANSWER KEY

1. p = q × 2a, unit: C·m
2. $V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$
3. $V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$
4. V = 0
5. Zero due to symmetry
6. $V \propto \frac{1}{r^2}$
7. $W = pE(\cos\theta_1 - \cos\theta_2)$
8. $U = -pE\cos\theta$
9. Stable: θ = 0°, Unstable: θ = 180°
10. Cosine curve

11. 8 × 10⁻⁸ C·m
12. 9 × 10³ V
13. 0
14. 5 × 10⁻² N·m
15. 0.3 J
16. −0.05 J
17. 2pE
18. ±10⁻³ J
19. 90°
20. $W = pE(1 - \cos\theta)$

✏️ DETAILED SOLUTIONS

Q11 Solution

Dipole moment:
p = q × d
= 2×10⁻⁶ × 0.04
= 8×10⁻⁸ C·m

Q12 Solution

Axial potential:

V = (1/4πε₀)(p / r²)

= 9×10⁹ × (4×10⁻⁸ / 0.2²)
= 9×10⁹ × (4×10⁻⁸ / 0.04)
= 9×10⁹ × 10⁻⁶
= 9×10³ V

Q13 Solution

On equatorial line:
V = 0 (symmetry cancels)

Q14 Solution

Torque:
τ = pE sinθ
= 2×10⁻⁶ × 5×10⁴ × sin30°
= 2×10⁻⁶ × 5×10⁴ × 0.5
= 5×10⁻² N·m

Q15 Solution

Work done:
W = pE(cos0° − cos90°)
= pE(1 − 0)
= 3×10⁻⁶ × 10⁵
= 0.3 J

Q16 Solution

Potential energy:
U = −pE cosθ
= −5×10⁻⁶ × 2×10⁴ × cos60°
= −5×10⁻⁶ × 2×10⁴ × 0.5
= −0.05 J

Q17 Solution

ΔU = U_final − U_initial

= [−pE cos180° − (−pE cos0°)]
= [pE − (−pE)]
= 2pE

Q18 Solution

Max: +pE
Min: −pE

= ± (10⁻⁸ × 10⁵)
= ±10⁻³ J

Q19 Solution

U = −pE cosθ = 0
⇒ cosθ = 0
⇒ θ = 90°

Q20 Solution

W = ΔU

= −pE cosθ − (−pE)
= pE(1 − cosθ)

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