If a dielectric slab is introduced between the plates of a parallel plate capacitor after the battery is disconnected, then how do the following quantities change. (i) Charge (ii) Potential (iii) Capacitance (iv) Energy

When a dielectric slab is inserted between the plates of a parallel plate capacitor after the battery has been disconnected, several important changes occur in the capacitor's properties. Let's break down how each of the following quantities—charge, potential, capacitance, and energy—are affected by this action.

Understanding the Impact on Charge

Initially, when the capacitor is charged while connected to a battery, it stores a certain amount of charge (Q). Once the battery is disconnected, the charge on the capacitor plates remains constant because there is no external circuit to allow charge to flow in or out. When the dielectric slab is introduced, the charge on the plates does not change.

Effects on Potential

The potential difference (V) across the capacitor is related to the charge and capacitance by the formula:

• V = Q / C

Since the charge (Q) remains constant and the capacitance (C) increases when the dielectric is added (due to the dielectric constant, κ), the potential difference across the capacitor decreases. This can be understood as follows: the dielectric material reduces the electric field between the plates, which in turn lowers the voltage.

Capacitance Changes

The capacitance of a parallel plate capacitor is given by:

• C = (κε₀A) / d

Here, κ is the dielectric constant, ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between them. When a dielectric slab is inserted, the dielectric constant (κ) increases the capacitance. Therefore, the new capacitance becomes:

• C' = κC

This increase in capacitance occurs because the dielectric material allows the capacitor to store more charge for the same potential difference.

Energy Considerations

The energy (U) stored in a capacitor can be expressed as:

• U = (1/2) CV²

Since the charge remains constant and the potential decreases, we can analyze how energy changes. The energy can also be expressed in terms of charge and capacitance:

• U = (Q²) / (2C)

As the capacitance increases (C' > C) while Q remains constant, the energy stored in the capacitor decreases. This is because the energy is inversely proportional to capacitance when charge is constant. Thus, the introduction of the dielectric slab results in a decrease in the energy stored in the capacitor.

Summary of Changes

• Charge: Remains constant.
• Potential: Decreases.
• Capacitance: Increases.
• Energy: Decreases.

In summary, inserting a dielectric slab into a disconnected parallel plate capacitor leads to an increase in capacitance, a decrease in potential, and a decrease in energy, while the charge remains unchanged. This illustrates the fascinating interplay between electric fields, materials, and energy storage in capacitors.