Gauss's Law Applications
Electrostatics
45
JEE Qs
20%
Hard
90
min
Always visualize the electric field lines and use the symmetry of the charge distribution to choose the simplest possible Gaussian surface for efficient calculation of the electric field.
🧮 Key Formulas
✅ Key Points for JEE
- 1Gauss's Law is universally true but practically useful for calculating electric fields ONLY for charge distributions with high degrees of symmetry (spherical, cylindrical, planar).
- 2The most crucial step is choosing the correct 'Gaussian surface' (an imaginary closed surface) such that the electric field is either constant and perpendicular to the surface, or parallel to the surface (E ⋅ dA = 0) over different parts.
- 3Carefully determine 'Q_enclosed', which is the net charge strictly *within* the Gaussian surface, considering volume, surface, or linear charge densities and any induced charges.
- 4For conductors in electrostatic equilibrium, the electric field *inside* the conductor is always zero, and any net charge resides entirely on its surface.
- 5The electric field just outside the surface of a conductor is always perpendicular to the surface and has a magnitude E = σ / ε₀, where σ is the local surface charge density.
⚠️ Common Mistakes
- ✕Incorrectly identifying the symmetry of the charge distribution and choosing an inappropriate Gaussian surface, leading to complex or unsolvable integrals.
- ✕Miscalculating the 'Q_enclosed' term, especially in problems involving multiple layers, hollow conductors, or varying charge densities.
- ✕Forgetting that the electric field inside a conductor in electrostatic equilibrium is zero, or incorrectly applying the E = σ/ε₀ formula inside a conductor.
- ✕Assuming E is constant over the entire Gaussian surface when only E ⋅ dA is constant or zero over specific parts due to symmetry.
📝 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