Important questions based on NCERT syllabus for Chapter 2 - Electrostatic Potential and Capacitance:
Question-1: Two charges 2 μC and –2 μC are placed at points A and B 6 cm apart
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
Q1 = 2 μC
Q2 = ‒2 μC
r = 6 cm
(a) Since both charges are equal and opposite, they will cancel out each other’s effect at the centre of line joining them, and the plane passing through it will have equal potential (i.e. zero).
(b) Normal to the plane in the direction AB.
Question-2: A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
C = 600 pF
V = 200 V
C’ = 600 pF
Initial energy, U = CV2 / 2 = 1.2 × 10^‒5 J
Since half of energy initially stored is lost in the form of heat and electromagnetic radiation, therefore
Energy lost = U / 2 = 6 x 10^‒6 J.
Question-3: A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
C = 12 pF
V = 50 V
Energy stored = CV2 / 2 = 1.5 × 10‒8 J