When two identical capacitors are connected in a circuit, the behavior of the system can change significantly when a dielectric slab is introduced into one of the capacitors. Let's break down the effects on capacitance, charge, potential difference, and stored energy in a clear and structured manner.
Understanding Capacitors and Dielectrics
A capacitor is a device that stores electrical energy in an electric field. The capacitance (C) of a capacitor is defined as the amount of charge (Q) it can store per unit voltage (V), expressed by the formula:
C = Q/V
When a dielectric material is inserted between the plates of a capacitor, it affects the capacitance. The capacitance increases by a factor known as the dielectric constant (κ), which is a property of the material. The new capacitance can be expressed as:
C' = κC
Capacitance Changes with Dielectric
In your scenario, when the dielectric slab is inserted into capacitor-2 while the battery remains connected, the capacitance of capacitor-2 increases. If we denote the original capacitance of both capacitors as C, then:
- Capacitance of capacitor-1 remains C.
- Capacitance of capacitor-2 becomes C' = κC.
Charge Distribution and Potential Difference
With the battery connected, it maintains a constant voltage (V) across the entire circuit. The charge on each capacitor can be calculated using the formula:
Q = CV
For capacitor-1, the charge remains:
Q1 = CV
For capacitor-2, since its capacitance has increased, the charge will also increase:
Q2 = C'V = (κC)V = κQ
This means that the charge on capacitor-2 increases due to the presence of the dielectric slab, while the charge on capacitor-1 remains unchanged.
Potential Difference Across Each Capacitor
The potential difference across each capacitor can be calculated using the formula:
V = Q/C
For capacitor-1, the potential difference remains:
V1 = Q1/C = V
For capacitor-2, since the charge has increased, the potential difference will change:
V2 = Q2/C' = (κQ)/(κC) = V
Interestingly, even though the charge on capacitor-2 has increased, the potential difference across it remains equal to the battery voltage due to the connection.
Stored Energy in Each Capacitor
The energy (U) stored in a capacitor can be calculated using the formula:
U = 1/2 CV^2
For capacitor-1, the energy stored is:
U1 = 1/2 C V^2
For capacitor-2, with the increased capacitance, the energy stored becomes:
U2 = 1/2 C' V^2 = 1/2 (κC) V^2 = κ(1/2 C V^2) = κU1
This indicates that the energy stored in capacitor-2 increases due to the dielectric slab, while the energy in capacitor-1 remains constant.
Summary of Changes
- Capacitance of capacitor-1 remains C; capacitance of capacitor-2 increases to C' = κC.
- Charge on capacitor-1 remains Q; charge on capacitor-2 increases to Q2 = κQ.
- Potential difference across both capacitors remains equal to the battery voltage (V).
- Stored energy in capacitor-1 remains U1; stored energy in capacitor-2 increases to U2 = κU1.
In conclusion, introducing a dielectric slab into one capacitor while keeping the battery connected leads to an increase in capacitance, charge, and stored energy in that capacitor, while the other capacitor's values remain unchanged. This illustrates the fascinating interplay between capacitance, charge, and energy in capacitors.
