Question

Read the following statements about a parallel-plate capacitor and identify the correct option: (I) The capacitance of a parallel-plate capacitor in vacuum is given by C = ε0A/d, where A is the area of each plate and d is the separation between them. (II) When a dielectric slab of dielectric constant K is fully inserted between the plates, the capacitance becomes K times its earlier value. (III) The energy stored per unit volume (energy density) in the electric field between the plates (in vacuum) is u = ½ ε0E². (IV) If the plates of an isolated charged capacitor are pulled farther apart at constant charge, the energy stored in the capacitor decreases.

Case study: A parallel-plate capacitor of capacitance 6 μF is charged by a battery of EMF 12 V. The battery is then disconnected, and a dielectric slab of dielectric constant K = 4 is fully introduced between the plates. The new potential difference across the capacitor and the new energy stored in it are respectively:

(a) A 900 pF capacitor is charged by 100 V battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic energy stored by the system?

A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?