Parallel Plate Capacitor 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 Numerical for Class-12 Nootan ISC Physics Ch-4 Capacitors and Dielectrics
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-4 | Capacitors and dielectrics |
| Topics | Numericals on Parallel Plate Capacitor |
| Academic Session | 2025-2026 |
Numericals on Parallel Plate Capacitor
Parallel Plate Capacitor Numerical for Class-12 Nootan ISC Physics Ch-4 Capacitors and Dielectrics
Que-3. A PD of 250 V exists across the plates of a 25 µF capacitor. Find the charge on each plate.
Ans-3 Potential difference = 250V
Capacitor = 25 µF
Q = CV
Q = 250 × 25 × 10^-6
= 6.25×10^-3
Charge on positive plates = 6.25×10^-3
Charge on negative plates = -6.25×10^-3
Que-4. What distance d between the two parallel plates, each of area A = 0.0314 m², of an air capacitor be kept so that its capacitance is same as of a conducting sphere of radius a = 0.5 m?
Ans-4 Capacitance on spherical conductor = Capacitance of parallel plate capacitor
= 4πε0 r = (ε0A)/d
r = radius of conductor
d = distance between plates.
d = (ε0A)/r x {1/(4πε0)}
d = A/4πr = 0.0134/(4 x 3.14 x 0.5)
= 0.5 x 10¯³m = 0.5 mm.
Que-5. An air parallel-plate capacitor has a plate area of 6 x 10^-3 m² and a plate separation of 3 mm. Calculate its capacitance. If this capacitor is connected to a 100 V supply. What is the charge on each plate of the capacitor?
Ans-5 Given : A = 6 x 10¯³ m²
d = 3mm = 3 x 10(ε0A)/m
V = 100V
C = (ε0A)/d
= {8.85 x 10¯¹² x 6 x 10¯³}/(3 x 10¯³)
= 17.70 x 10¯¹² = 18 pF
Q = CV
= 18 x 10¯¹² x 100
= 1.8 x 10^-9 C.
— : End Parallel Plate Capacitor Numerical for Class-12 Nootan ISC Physics Ch-4 Capacitors and Dielectrics :–
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