Capacitors — Combinations & Energy
Electrostatics
88
JEE Qs
22%
Hard
100
min
Master charge conservation for isolated capacitor plates and understand the origin and calculation of energy loss during charge redistribution for complex problems.
🧮 Key Formulas
✅ Key Points for JEE
- 1For series capacitors, the charge (Q) on each capacitor is the same, and the total voltage is the sum of individual voltages. For parallel capacitors, the voltage (V) across each is the same, and the total charge is the sum of individual charges.
- 2The rules for finding equivalent capacitance for series and parallel combinations are *opposite* to those for resistors.
- 3In problems involving connecting capacitors (especially isolated ones), charge is conserved on isolated plates or sections of the circuit, leading to a common potential after redistribution.
- 4Energy is always lost (dissipated as heat or electromagnetic radiation) during charge redistribution when two capacitors at different potentials are connected, or when a charged capacitor is connected to an uncharged one.
- 5Symmetry principles (e.g., Wheatstone bridge symmetry, folding symmetry) can significantly simplify complex capacitor networks to find equivalent capacitance.
⚠️ Common Mistakes
- ✕Confusing the series/parallel combination rules for capacitors with those for resistors (e.g., adding reciprocals for parallel capacitors).
- ✕Failing to correctly apply the principle of charge conservation, especially when dealing with isolated plates or portions of a circuit during connection/re-connection scenarios.
- ✕Incorrectly calculating energy loss during charge redistribution by simply comparing initial and final total stored energy, without accounting for the energy dissipated as heat.
- ✕Assuming total energy of the system is conserved in all capacitor connection problems, neglecting the energy converted to heat or EM radiation.
📝 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 I, Chapter 2: Electrostatic Potential and Capacitance