Free NEET Physics multiple-choice questions on Capacitance & Dielectrics. Attempt each question and reveal the answer with a full explanation.
A parallel plate capacitor is charged and then the battery is disconnected. If a dielectric slab is inserted between the plates, which of the following remains constant? Charge Potential difference Energy stored Capacitance If the potential of a capacitor having capacity 6 F is increased from 10 V to 20 V , then the increase in its energy will be: 9 10 -4 J 12 10 -6 J 4 10 -4 J 3 10 -4 J The capacitance of a parallel plate capacitor is C . If the separation between the plates is doubled and a dielectric medium of constant K=3 is filled between them, the new capacitance is: 1.5C 3C 6C 0.5C A capacitor of 4 F is connected in series with a 6 F capacitor. A potential of 500 V is applied to the outer plates. The charge on each capacitor is: 1200 C 2500 C 1000 C 500 C A parallel plate capacitor having a capacitance of 12 pF is charged by a battery to a potential difference of 10 V between its plates. The energy stored in the capacitor is: 6 10 -10 J 12 10 -10 J 6 10 -12 J 1.2 10 -10 J The displacement current flows in the dielectric of a capacitor when the potential difference between its plates: is changing with time is constant becomes zero is very high A capacitor of capacitance C 1 is charged to a potential V 0 and then connected in parallel to an uncharged capacitor of capacitance C 2 . The common potential V of the combination will be: C 1 V 0 C 1 + C 2 C 2 V 0 C 1 + C 2 (C 1 + C 2)V 0 C 1 V 0 The capacitance of a spherical conductor is 1 F . Its radius is approximately: 9 km 1.1 km 1 m 100 m What is the net charge on a charged capacitor? Zero Q 2Q Q/2 The unit of electric susceptibility is: Unitless C/m² N/C V/m When a capacitor is charged, the energy is stored in: The electric field between the plates The magnetic field between the plates The wires of the circuit The plates themselves Three capacitors of capacitances 2 F , 3 F , and 6 F are connected in series. The equivalent capacitance is: 1 F 11 F 0.5 F 2 F If the distance between the plates of a parallel plate capacitor is halved and the area is doubled, the capacitance: Increases 4 times Decreases 4 times Remains same Increases 2 times Three capacitors of 2 F, 3 F and 5 F are connected in parallel. The total capacitance is: 10 F 1 F 30/31 F 0.5 F Dielectric constant for a metal is: Infinite Zero 1 -1 The force of attraction between the plates of a parallel plate capacitor is: Q 2 / (2ε 0 A) Q 2 / (ε 0 A) Q / (2ε 0 A) Q 2 / (2ε 0 A 2) Four capacitors are connected as shown in the figure. The equivalent capacitance between A and B is ( C 1 = 3 F, C 2 = 2 F, C 3 = 6 F, C 4 = 4 F ): 5 F 15 F 2 F 10 F A parallel plate capacitor is charged by a battery and then disconnected. If the distance between the plates is doubled, the energy stored in the capacitor will: be doubled be halved remain same become four times The capacitance of a parallel plate capacitor is 10 F . If a dielectric slab of K=2 is inserted to fill half the volume as shown in figure, the new capacitance will be: 15 F 20 F 5 F 12 F A parallel plate capacitor of capacitance 20 F is being charged by a voltage source whose potential is changing at the rate of 3 V/s . The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively: 60 A, 60 A 60 A, 0 0, 60 A 30 A, 30 A The work done in placing a charge of 8 10 -18 C on a condenser of capacity 100 F is: 3.2 10 -31 J 1.6 10 -32 J 3.1 10 -26 J 4 10 -10 J A parallel plate capacitor is connected to a battery. A dielectric slab is then inserted between the plates. In this process: The charge on the plates increases The potential difference increases The capacitance decreases The stored energy decreases The capacitance of a spherical conductor of radius R is proportional to: R R 2 1/R R 1/2 A capacitor of 10 F is charged to 100 V . It is then connected to another capacitor of 5 F charged to 50 V with like plates together. The common potential is: 83.3 V 75 V 50 V 100 V A parallel plate capacitor is charged and then disconnected from the battery. If the distance between the plates is increased, the energy stored in the capacitor: Increases Decreases Remains constant First increases then decreases A dielectric slab of constant K is inserted into a charged capacitor (battery disconnected) covering only half the area. The new capacitance is: C(1+K)/2 C(1+K) 2CK/(1+K) CK Two identical capacitors C 1 and C 2 of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor C 1 using battery of emf V . Now disconnecting a and b the terminals b and c are connected. What is the percentage loss of energy? 50% 0% 75% 25% A capacitor is charged by a battery and then disconnected. A dielectric slab is now inserted between the plates. Which of the following statements is correct? The potential difference decreases and the energy stored decreases The potential difference increases and the energy stored increases The potential difference decreases and the energy stored increases The potential difference remains constant and the energy stored decreases A parallel plate capacitor is filled with two dielectrics of dielectric constants K 1 and K 2 in series, each of thickness d/2 . The equivalent dielectric constant K of the capacitor is: 2K 1 K 2 K 1 + K 2 K 1 + K 2 2 K 1 + K 2 K 1 K 2 A parallel plate capacitor is charged and then disconnected from the battery. A dielectric slab is now introduced between the plates. This results in: Decrease in potential difference and increase in capacitance Increase in potential difference and decrease in capacitance No change in potential and capacitance Decrease in both potential and capacitance A metal plate of thickness d/2 is inserted between the plates of a parallel plate capacitor of plate separation d and capacitance C . The new capacitance is: 2C C/2 4C 2 C Two capacitors 3 F and 6 F are connected in series. A potential difference of 900 V is applied across the combination. The potential difference across the 3 F capacitor is: 600 V 300 V 450 V 900 V The capacitance of a parallel plate capacitor with air as medium is 6 μ F . With the introduction of a dielectric medium, the capacitance becomes 30 μ F . The permittivity of the medium is: ( ε 0 = 8.85 10 -12 C 2 N -1 M -2 ) 0.44 10 -10 C 2 N -1 M -2 5.00 C 2 N -1 M -2 0.44 10 -13 C 2 N -1 M -2 1.77 10 -12 C 2 N -1 M -2 A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and the area of each plate is A , the energy stored in the capacitor is: 1 2 ε 0 E 2 Ad E 2 Ad ε 0 1 2 ε 0 E 2 ε 0 EAd Two thin dielectric slabs of dielectric constants K 1 and K 2 ( K 1 < K 2 ) are inserted between the plates of a parallel plate capacitor, as shown in the figure. The variation of electric field E with distance x as measured from plate P is correctly given by: The field E is highest in air gaps and lowest in slab K 2 . The field E is constant everywhere. The field E is highest in slab K 2 . The field E is zero inside the slabs. A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles: The voltage across the plates increases The capacitance increases The charge on the capacitor increases The electrostatic energy stored decreases Two identical capacitors are connected in parallel and charged to a potential V . They are then separated and connected in series (positive plate of one to negative of other). The potential difference between the free plates is: 2V V V/2 Zero A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B . Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF . The electrostatic energy stored by the system is: 2.25 10 -6 J 4.5 10 -6 J 1.125 10 -6 J 9 10 -6 J A capacitor of 2 μ F is charged as shown in the circuit. When the switch S is turned to position 2, the percentage of its stored energy dissipated is: 80 % 0 % 20 % 75 % Two capacitors of capacitance 6 μ F and 3 μ F are connected in series and a potential difference of 18 V is applied across the combination. They are then disconnected and reconnected in parallel. The potential difference across the parallel combination is: 8 V 12 V 6 V 15 V A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system: Decreases by a factor of 2 Increases by a factor of 2 Remains the same Decreases by a factor of 4 The capacitance of a parallel plate capacitor is C with air between the plates. When a dielectric slab of dielectric constant K and thickness 3d/4 (where d is plate separation) is inserted between the plates, the new capacitance C' will be: 4KC (K+3) C 4 (K+3) C K+3 4K C 3K K+4 C If a slab of insulating material 4 10 -5 m thick is introduced between the plates of a parallel plate capacitor, the distance between the plates has to be increased by 3.5 10 -5 m to restore the capacity to original value. The dielectric constant of the material is: 8 6 10 12 A parallel plate capacitor is charged to a potential V . The battery is then disconnected and a dielectric slab of constant K is inserted to fill the space. If the initial capacitance was C , the work done by an external agent in inserting the slab is: CV 2 2 ( 1 K - 1) CV 2 2 (K - 1) CV 2 2K Zero A parallel plate capacitor with plate area A and separation d has capacitance C . If the area is doubled and separation is halved, the new capacitance is: 4C 2C C C/4 Three identical capacitors, P , Q and S , each of the capacitance C , are connected to a battery of voltage V , as shown in the figure. If the energy stored in the capacitor P and total energy stored in the system are U P and U T , respectively, then the ratio U P U T is: 2/3 1/3 1/2 1/6 A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of same area but thickness d/2 is inserted between the plates as shown in figure. The capacitance of the capacitor will: increase decrease remain same become zero Two parallel metal plates having charges +Q and -Q face each other at a certain distance. A dielectric slab is introduced between them. The electric field between the plates: decreases increases remains same becomes zero A parallel plate capacitor is charged to a potential difference V and then disconnected. A slab of dielectric constant K is now inserted between the plates. The new potential difference is: V/K KV V V/K 2 Three capacitors each of capacitance C and breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be: C 3 , 3V 3C, V 3 C 3 , V 3 3C, 3V