Parallel Plate Capacitor with Dielectric Slab between Plates Numerical for Class-12 Nootan ISC Physics Ch-4 Capacitors and Dielectrics. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Parallel Plate Capacitor with Dielectric Slab between Plates Numerical for Class-12 Nootan ISC Physics Ch-4
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-4 | Capacitors and dielectrics |
| Topics | Numericals on Parallel Plate Capacitor with Dielectric Slab between Plates |
| Academic Session | 2025-2026 |
Numericals on Parallel Plate Capacitor with Dielectric Slab between Plates
Class-12 Nootan ISC Physics Ch-4 Capacitors and Dielectrics.
Que-6. The area of each plate of a parallel-plate capacitor is 100 cm² and the distance between them is 0.05 cm. When it is filled with a dielectric, its capacitance becomes 3.54 x 10^-10 F. Find the dielectric constant.
Ans-6 Capacitance of capacitor with dielectric
C = kε0A/d
=> k = Cd/ε0A
= 3.54 x 10^-10 x 0.05 x 10^-2 / 8.85 x 10^-12 x 100 x 10^-4
= 3.54 x 5 / 8.85
= 2
Que-7. How much area of paper will be required to construct a parallel-plate capacitor of 0.004 µF, if the dielectric constant of the paper be 2.5 and its thickness be 0.025 mm?
Ans-7 C = kε0A/d
A = Cd/kε0
=> 0.004 x 10^-6 x 0.025 x 10^-3 / 2.5 x 8.85 x 10^-12
=> 100 x 10^-12 x 10^-3 / 2.5 x 8.85 x 10^-12
=> 4.52 x 10^-3 m
Que-8. We have a glass slab (K = 7) 4.0 mm thick, a mica foil (K = 6) 0.20 mm thick and an amber plate (K = 2) 2.0 cm thick. Which one be placed between the plates of a parallel-plate capacitor to obtain maximum capacitance?
Ans-8 C = kε0A/d = (k/d) ε0A
=> C ∝ k/d
for mica foil k/d = 6/(0.2 x 10^-3) = 3 x 10^4 (max)
∴ Mica foil is better
Que-9. The potential difference between the plates of a parallel- plate capacitor is 200 V. The area of each plate is 100 cm² and the distance between them is 1 mm. If the medium between them is air, then calculate the charge taken by the capacitor. If there is a medium of dielectric constant 2.5 between the plates, then what will be the potential difference for the same charge?
Ans-9 V = 200 V A = 100 x 10^-4 m^2
d = 1 x 10^-3 m
Q = CV = ε0A/d x V
=> 8.85 x 100 x 10^-4 x 200 / 1 x 10^-3
=> 1.76 x 10^-8 C
Potential with dielectric constant
= 200/2.5 = 80 V (V’ = V/k)
Que-10. A paper is placed between two plates of copper, area of each 100 cm². The thickness of the paper is 0.005 cm and its dielectric constant is 2.5. If the paper can bear an electric field of 5 x 10^5 V/cm, then calculate the maximum voltage up to which the capacitor can be charged. How much charge will be stored on the capacitor?
Ans-10 E = V/d => V = E.d
E = 5 x 10^5 V/cm = 5 x 10^5 x (10^-2) m^-1
=> 5 x 10^7 volt/m
∴ V = 5 x 10^7 x 0.005 x 10^-2
=> 2500 V
again charge Q = CV
=> kε0A/d x V = ε0A.E
=> 2.5 x 8.85 x 10^-12 x 100 x 10^-4 x 5 x 10^7
=> 1.1 x 10^-5 C
Que-11. When a 20 V battery is connected to an air-capacitor, a charge of 30 µC is obtained on the capacitor. When oil is filled between the plates of the capacitor, then a charge of 75 µC is obtained. Calculate (i) the dielectric constant of the oil, (ii) the energy contained in the oil-capacitor.
Ans-11 Charge in capacitor with dielectric
=> k.charge in air capacitor
=> 75 µC = k x 30 µC
=> k = 75/30 = 2.5
again energy = 1/2 CV^2 = 1/2 (CV).V = 1/2 q.V
=> 1/2 x 75 x 10^-6 x 20
=> 7.5 x 10^-4 J
Que-12. (a) A parallel-plate air capacitor has a plate area of 6 x 10^-3 m² and a plate separation of 3 mm. A PD of 100 V is established between its plates by a battery. After disconnecting the battery, the space between the plates is filled by a 3 mm thick mica sheet of dielectric constant K = 6. Find (i) the new PD between the plates, (ii) the final capacitance and (iii) the final charge on the capacitor.
Ans-12 (a) When di electric slab is inserted after removing battery then
(i) V = V0/k = 100/6 volt
(ii) C = KC0 = kε0A/d
=> 6 x 8.85 x 10^-12 x 6 x 10^-3 / 3 x 10^-3
=> 108 x 10^-12 F = 108 pF
(iii) Q = CV
=> 108 x 10^-12 x 100/6
=> 1.8 x 10^-9 C
(b) if battery is connected then potential will be same but charge will increase
∴ V = 100 V
Q = 1.8 x 10^-9 x 6
=> 1.08 x 10^-8 C
but capacitance will be same
i.e. 108 pF
— : End Parallel Plate Capacitor Numerical for Class-12 Nootan ISC Physics Ch-4 Capacitors and Dielectrics :–
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