Electric Dipole — Torque, potential, field
Electrostatics
18
JEE Qs
8%
Hard
75
min
Master the vector nature of electric dipole moment, field, torque, and potential energy, paying close attention to directions and signs in problem-solving.
🧮 Key Formulas
✅ Key Points for JEE
- 1Electric dipole moment (p) is a vector from the negative charge to the positive charge.
- 2For a short dipole, electric field falls off as 1/r^3 and electric potential falls off as 1/r^2, unlike point charges (1/r^2 for E, 1/r for V).
- 3Torque on a dipole in a uniform electric field tends to align the dipole moment (p) with the electric field (E).
- 4The potential energy of a dipole is minimum (most stable equilibrium) when p is parallel to E (θ=0°) and maximum (unstable equilibrium) when p is anti-parallel to E (θ=180°).
- 5Net force on an electric dipole in a *uniform* electric field is zero, but in a *non-uniform* field, it experiences a net force in addition to torque.
⚠️ Common Mistakes
- ✕Incorrectly defining the direction of the dipole moment vector (p).
- ✕Using r^2 or r for field/potential decay instead of r^3 and r^2, respectively, for a short dipole.
- ✕Errors in sign conventions for potential energy calculations (U = -pE cosθ) or when calculating work done by/against the field.
- ✕Confusing the axial and equatorial field/potential directions and magnitudes.
📝 Practice Questions
See allQ38.A capacitor, C1 = 6μ F is charged to a potential difference of V0 = 5 V using a 5 V battery. The battery is removed and another capacitor, C2 = 12μ F is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the 2025 (29 Jan Shift 2) JEE Main Previous Year Paper charges (q1 and q2) on the capacitors C1 and C2 when equilibrium condition is reached. (1) q1 = 10μC, q2 = 20μC (2) q1 = 30μC, q2 = 15μC (3) q1 = 20μC, q2 = 10μC (4) q1 = 15μC, q2 = 30μC
Q44.A point charge causes an electric flux of −2 × 104Nm2C−1 to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is : (Given ϵ0 = 8.85 × 10−12C2 N−1 m−2 ) (1) 15.7 × 10−8C (2) 17.7 × 10−8C (3) −15.7 × 10−8C (4) −17.7 × 10−8C
Q45.An electric dipole is placed at a distance of 2 cm from an infinite plane sheet having positive charge density σo . Choose the correct option from the following. (1) Potential energy and torque both are maximum. (2) Torque on dipole is zero and net force is directed away from the sheet. (3) Torque on dipole is zero and net force acts (4) Potential energy of dipole is minimum and torque towards the sheet. is zero.
Q26.An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity 106 m/s. If the magnitude of the electric field between the plates is 9.1 V/cm , then the vertical component of velocity of electron is (mass of electron = 9.1 × 10−31 kg and charge of electron = 1.6 × 10−19C ) (1) 0 (2) 1 × 106 m/s (3) 16 × 106 m/s (4) 16 × 104 m/s
Q38.A line charge of length ' a ' is kept at the center of an edge BC of a cube ABCDEFGH having edge length ' 2 a ' as shown in the figure. If the density of line charge is λ C per unit length, then the total electric flux through all the faces of the cube will be . (Take, ϵ0 as the free space permittivity) (1) λa (2) λa 2ϵ0 4ϵ0 (3) λa (4) λa 16ϵ0 8ϵ0
Q44.A parallel-plate capacitor of capacitance 40μ F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are (1) 4 mC and 0.2 J (2) 8 mC and 2.0 J (3) 2 mC and 0.4 J (4) 2 mC and 0.2 J 2025 (22 Jan Shift 1) JEE Main Previous Year Paper
NCERT Chapters
- Class 12 Physics Part 1 Ch 1: Electric Charges and Fields