Problem Statement:
A parallel plate capacitor is connected to a battery, and the battery remains connected while a dielectric slab is inserted between the plates. We are asked to explain the changes, if any, in the following quantities:
1. Potential difference between the plates.
2. Electric field between the plates.
3. Energy stored in the capacitor.
Key Concepts:
• Capacitance of a Parallel Plate Capacitor: The capacitance CC of a parallel plate capacitor without a dielectric is given by:
C0=ε0AdC_0 = \frac{\varepsilon_0 A}{d}
Where:
o ε0\varepsilon_0 is the permittivity of free space,
o AA is the area of the plates,
o dd is the separation between the plates.
When a dielectric material with dielectric constant κ\kappa is inserted between the plates, the capacitance increases to:
C=κC0=κε0AdC = \kappa C_0 = \frac{\kappa \varepsilon_0 A}{d}
Where:
o κ\kappa is the dielectric constant of the material.
• Potential Difference: The potential difference VV across the plates of a capacitor is related to the charge QQ and capacitance CC by:
V=QCV = \frac{Q}{C}
Since the battery remains connected, the charge on the capacitor can adjust to maintain the constant potential difference provided by the battery.
• Electric Field: The electric field EE between the plates of a capacitor without a dielectric is:
E0=VdE_0 = \frac{V}{d}
When a dielectric is inserted, the electric field reduces because the dielectric material polarizes and reduces the effective electric field. The new electric field becomes:
E=E0κE = \frac{E_0}{\kappa}
So, the electric field is reduced by a factor of κ\kappa.
• Energy Stored in the Capacitor: The energy UU stored in a capacitor is given by:
U=12CV2U = \frac{1}{2} C V^2
Since the battery is connected, the potential difference VV remains constant. However, as the capacitance increases due to the dielectric, the energy stored in the capacitor increases. The new energy stored is:
U′=12C′V2U' = \frac{1}{2} C' V^2
Where C′C' is the capacitance after the dielectric is inserted. Since C′C' is larger than C0C_0, the energy stored increases.
Changes After Inserting the Dielectric Slab:
1. Potential Difference Between the Plates:
o Since the battery remains connected, the potential difference VV across the plates will remain the same. The battery maintains the constant voltage across the capacitor.
2. Electric Field Between the Plates:
o The electric field between the plates will decrease because the dielectric slab reduces the effective electric field. The new electric field will be: E=E0κE = \frac{E_0}{\kappa}
This decrease is due to the polarization of the dielectric material, which opposes the electric field.
3. Energy Stored in the Capacitor:
o The energy stored in the capacitor will increase because the capacitance increases when the dielectric is inserted. Since the voltage remains constant (due to the battery), the increase in capacitance results in more energy being stored: U=12CV2U = \frac{1}{2} C V^2
Therefore, with a larger capacitance CC, the stored energy increases.
• Potential difference between the plates: Remains the same, as the battery keeps the voltage constant.
• Electric field between the plates: Decreases because of the dielectric insertion.
• Energy stored in the capacitor: Increases due to the increase in capacitance with the dielectric inserted.
