A parallel-plate capacitor with plate area A = 2.0 m² and plate separation d = 3.0 mm is connected to a 45-V battery (Fig. 24–39a). With the capacitor still connected to the battery, a slab of plastic with dielectric constant K = 3.2 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric. What are the new values of charge, electric field, capacitance, and the energy U stored in the capacitor?

Verified step by step guidance

1. Step 1: Begin by recalling the formula for the capacitance of a parallel-plate capacitor without a dielectric: \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the plate area, and \( d \) is the plate separation. Substitute the given values \( A = 2.0 \, \text{m}^2 \) and \( d = 3.0 \, \text{mm} = 3.0 \times 10^{-3} \, \text{m} \).
2. Step 2: When the dielectric is introduced, the capacitance increases by a factor of the dielectric constant \( K \). The new capacitance is given by \( C' = K \cdot C \). Use \( K = 3.2 \) to calculate the new capacitance.
3. Step 3: The charge stored on the capacitor is given by \( Q = C \cdot V \), where \( V \) is the voltage across the capacitor. Since the capacitor remains connected to the battery, \( V \) remains constant at 45 V. Use the new capacitance \( C' \) to calculate the new charge \( Q' \).
4. Step 4: The electric field between the plates is given by \( E = \frac{V}{d} \). Since the plate separation \( d \) and voltage \( V \) remain unchanged, the electric field \( E \) remains the same even after the dielectric is introduced.
5. Step 5: The energy stored in the capacitor is given by \( U = \frac{1}{2} C V^2 \). Use the new capacitance \( C' \) and the constant voltage \( V = 45 \, \text{V} \) to calculate the new stored energy \( U' \).

Key Concepts

Capacitance

Capacitance is the ability of a capacitor to store charge per unit voltage, defined by the formula C = εA/d, where ε is the permittivity of the dielectric material, A is the plate area, and d is the separation between the plates. When a dielectric material is introduced, the capacitance increases by a factor equal to the dielectric constant (K), allowing the capacitor to store more charge at the same voltage.

Electric Field

The electric field (E) between the plates of a capacitor is defined as the voltage (V) across the plates divided by the separation (d), given by E = V/d. When a dielectric is inserted, the electric field decreases because the dielectric reduces the effective voltage across the plates, which is crucial for understanding how the capacitor behaves when the dielectric is present.

Energy Stored in a Capacitor

The energy (U) stored in a capacitor can be calculated using the formula U = 1/2 CV², where C is the capacitance and V is the voltage. When a dielectric is added, the capacitance increases, which in turn increases the energy stored in the capacitor, assuming the voltage remains constant. This relationship is essential for determining how the energy changes when the dielectric is introduced.