Electric Field — Due to point charge, dipole, ring, disc
Electrostatics
18
JEE Qs
8%
Hard
120
min
Master vector addition and integral calculus for continuous charge distributions, as they are fundamental to solving complex electric field problems.
🧮 Key Formulas
✅ Key Points for JEE
- 1Electric field is a vector quantity; always determine both magnitude and direction, then apply the superposition principle using vector addition.
- 2Symmetry is crucial: for extended charge distributions like rings and discs, exploit symmetry to determine which components of the field cancel out.
- 3Pay close attention to approximations (e.g., r >> a for dipoles, x >> R or x << R for ring/disc) as they lead to simplified and frequently tested limiting cases.
- 4Understand the distinction between electric field (force per unit charge) and electric force (actual force on a charge) and know when to apply each.
- 5For continuous charge distributions, correctly set up the integral by choosing an appropriate differential element (dQ) and expressing it in terms of charge density and geometry.
⚠️ Common Mistakes
- ✕Treating electric field as a scalar quantity and simply adding magnitudes without considering directions or vector components.
- ✕Errors in applying the superposition principle, especially when dealing with charges of different signs or multiple dipoles.
- ✕Confusing the formulas for axial and equatorial positions of a dipole, or misapplying limiting approximations for ring/disc fields.
- ✕Incorrectly setting up or solving the definite integral required for continuous charge distributions, often due to wrong differential element or limits.
📝 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 Ch 1: Electric Charges and Fields