⚡ ELECTRIC DIPOLE WORKSHEET (JEE / NEET LEVEL) QUIZ1

📘 SECTION A: THEORY / CONCEPTUAL (Q1–Q10)

Q1. An electric dipole consists of:

A. Two equal positive charges
B. Two equal negative charges
C. Two equal and opposite charges separated by a small distance
D. Two unequal charges

Q2. Dipole moment is defined as:

A. Product of charge and distance
B. Product of charge and square of distance
C. Ratio of charge and distance
D. Product of force and distance

Q3. Direction of dipole moment is:

A. From negative to positive charge
B. From positive to negative charge
C. Perpendicular to dipole axis
D. Random

Q4. Electric field on axial line of a dipole is:

A. Zero
B. Along dipole moment
C. Opposite to dipole moment
D. Perpendicular to dipole moment

Q5. Electric field on equatorial line of a dipole is:

A. Along dipole moment
B. Opposite to dipole moment
C. Perpendicular to dipole moment
D. Zero

Q6. Ratio of electric field at axial point to equatorial point (far field) is:

A. 1 : 1
B. 2 : 1
C. 1 : 2
D. 4 : 1

Q7. Torque on dipole in uniform electric field is maximum when:

A. θ = 0°
B. θ = 90°
C. θ = 180°
D. θ = 45°

Q8. Stable equilibrium of dipole occurs when:

A. Dipole parallel to electric field
B. Dipole perpendicular to electric field
C. Dipole anti-parallel
D. None

Q9. Potential energy of dipole in electric field is minimum when:

A. θ = 0°
B. θ = 90°
C. θ = 180°
D. Undefined

Q10. Work done to rotate dipole from 0° to 180° is:

A. pE
B. 2pE
C. -pE
D. Zero

🔢 SECTION B: NUMERICAL (Q11–Q20)

Q11. Dipole moment of charges ±2 μC separated by 3 cm is:

A. 6 × 10⁻⁸ C·m
B. 6 × 10⁻⁶ C·m
C. 3 × 10⁻⁸ C·m
D. 12 × 10⁻⁸ C·m

Q12. Electric field on axial line at distance r is:

$E = \frac{1}{4\pi\epsilon_0}\frac{2p}{r^3}$

If p = 10⁻⁶ C·m, r = 0.1 m, find E.

A. 1.8 × 10⁶ N/C
B. 1.8 × 10⁷ N/C
C. 9 × 10⁶ N/C
D. 3.6 × 10⁶ N/C

Q13. Electric field on equatorial line is:

$E = \frac{1}{4\pi\epsilon_0}\frac{p}{r^3}$

For same values as Q12, find E.

A. 9 × 10⁶ N/C
B. 4.5 × 10⁶ N/C
C. 1.8 × 10⁶ N/C
D. 3.6 × 10⁶ N/C

Q14. Torque on dipole:

$\tau = pE\sin\theta$

If p = 2 C·m, E = 5 N/C, θ = 90°, find torque.

A. 5 Nm
B. 10 Nm
C. 2 Nm
D. 0

Q15. Potential energy:

$U = -pE\cos\theta$

For p = 2, E = 5, θ = 0°, find U.

A. -10 J
B. 10 J
C. 5 J
D. 0

Q16. Work done from 0° to 90°:

A. pE
B. pE/2
C. Zero
D. -pE

Q17. If dipole is perpendicular to field, torque is:

A. Maximum
B. Minimum
C. Zero
D. Constant

Q18. Two charges ±q separated by 2a. Dipole moment:

A. qa
B. 2qa
C. q/2a
D. qa²

Q19. If distance doubles, electric field on axial line becomes:

A. Half
B. One-fourth
C. One-eighth
D. Same

Q20. If p doubles, field becomes:

A. Same
B. Double
C. Half
D. Quadruple

✅ ANSWER KEY

Q1–Q10:
1-C, 2-A, 3-A, 4-B, 5-B, 6-B, 7-B, 8-A, 9-A, 10-B

Q11–Q20:
11-A, 12-B, 13-A, 14-B, 15-A, 16-A, 17-A, 18-B, 19-C, 20-B

🧠 DETAILED SOLUTIONS

Q11 Solution

p = q × d = (2 × 10⁻⁶)(0.03) = 6 × 10⁻⁸ C·m

Q12 Solution

Substitute values → E = 9×10⁹ × (2×10⁻⁶)/(0.1³)
= 1.8 × 10⁷ N/C

Q13 Solution

E = 9×10⁹ × (10⁻⁶)/(0.1³) = 9 × 10⁶ N/C

Q14 Solution

τ = pE sin90° = 2 × 5 = 10 Nm

Q15 Solution

U = -pE cos0 = -2 × 5 = -10 J

Q16 Solution

W = pE (cos0 − cos90) = pE (1 − 0) = pE

Q17 Solution

Torque is maximum at 90°

Q18 Solution

p = q × (2a) = 2qa

Q19 Solution

E ∝ 1/r³ → doubling r → field becomes 1/8

Q20 Solution

E ∝ p → doubling p → field doubles

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