Electrostatic Potential and Capacitance Class 12 |NCERT Chapter 2 (Part 2)| CBSE NEET JEE| One Shot

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Timestamps:
0:00 Introduction
0:46 Capacitor
6:33 Behaviour of Conductors in Electric fields
15:02 Electrostatic Shielding
17:09 Electrical Capacitance
19:50 How does a capacitor work?
24:31 Types of Capacitors
26:06 Parallel plate capacitor
28:48 What is a Dielectric?
31:22 Polarisation of a Dielectric
34:22 Dielectrics & Polarisation
42:18 Parallel plate capacitor:Conducting slab
43:24 Capacitance:Parallel plate capacitor
46:54 Parallel plate capacitor:Dielectric slab
51:46 Problem 1
55:18 Problem 2
59:12 Combination of capacitors
1:01:31Capacitors in Series
1:03:18 Capacitors in Parallel
1:04:33 Problem 1
1:07:12 Problem 2
1:08:41 Problem 3
1:10:15 Problem 4
1:11:39 Problem 5
1:14:02 Energy stored in Capacitor
1:18:15 Energy density in a capacitor
1:19:32 Problem 1
In this video we will cover:-
1.A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?
2.A dielectric slab of thickness 1cm and dielectric constant 5 is placed between the plates of a parallel plate capacitor of plate area 0.01m2 and separation 2cm. Calculate the change in capacity on introduction of dielectric. What would be the change, if the dielectric slab were conducting?
3.To find the equivalent Capacitance of the given circuit?
4.To find the equivalent Capacitance of the given circuit?
5.To find the equivalent Capacitance of the given circuit?
6.To find the equivalent Capacitance of the given circuit?
7.Find the Equivalent capacitance between A & B. Given C1 = 3µF, C2 = C3 = C4 = 9µF.
8.A parallel plate 100 µF capacitor is charged to 500V, If the distance between its plates is halved, what will be the new potential difference between the plates and what will be the change in stored energy?
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