Dielectrics — Polarization, dielectric constant
Electrostatics
18
JEE Qs
8%
Hard
75
min
Master the application of dielectrics in capacitors for both constant charge (battery disconnected) and constant potential (battery connected) scenarios, as this is a frequent JEE test point.
🧮 Key Formulas
✅ Key Points for JEE
- 1Dielectric materials, when placed in an external electric field, polarize to create an internal induced electric field that opposes the external field, thus reducing the net electric field inside the material.
- 2The dielectric constant (K or ε_r) quantifies this reduction: E_net = E_0 / K. This directly implies V_net = V_0 / K across the dielectric if the field is uniform.
- 3For a capacitor, inserting a dielectric of constant K completely filling the space increases its capacitance by a factor of K: C = K C_0.
- 4Always distinguish between 'free charges' (on capacitor plates, responsible for E_0) and 'bound charges' (induced on dielectric surfaces due to polarization). The electric displacement vector D (D = ε_0 E + P) is independent of bound charges and simplifies Gauss's Law in the presence of dielectrics.
- 5Analyze problems involving dielectrics in capacitors by carefully considering two critical scenarios: (a) the battery is disconnected (charge Q remains constant) and (b) the battery remains connected (potential V remains constant). Each scenario leads to different changes in E, C, V, Q, and energy.
⚠️ Common Mistakes
- ✕Confusing the effects on charge, potential, electric field, and stored energy when a dielectric is introduced into a capacitor, especially when distinguishing between battery-connected and battery-disconnected cases.
- ✕Incorrectly applying Gauss's Law in the presence of dielectrics by failing to account for bound charges or by not using the electric displacement vector D appropriately.
- ✕Forgetting that the dielectric constant K is always greater than or equal to 1 (K=1 for vacuum/air) and thus always increases capacitance.
- ✕Not understanding that polarization creates an induced field *opposing* the external field, leading to a *reduction* in the net field inside the dielectric.
📝 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
- Class 12 Physics Part 1 Ch 2: Electrostatic Potential and Capacitance