AP Physics C E&M- 10.4 Dielectrics- Study Notes- New Syllabus
AP Physics C E&M- 10.4 Dielectrics – Study Notes
AP Physics C E&M- 10.4 Dielectrics – Study Notes – per latest Syllabus.
Key Concepts:
- Dielectrics and Capacitors
Dielectrics and Capacitors
A dielectric is an insulating material placed between the plates of a capacitor. It reduces the effective electric field within the capacitor and increases the capacitance by allowing more charge to be stored for the same potential difference. 
Effect of a Dielectric:
When a dielectric is inserted, its molecules become polarized (dipoles align with the external field).
- This polarization produces an induced electric field that opposes the original field between the plates.
- The net electric field is therefore reduced:
\( \mathrm{E_{net} = \dfrac{E_0}{\kappa}} \)
- \(\mathrm{E_0}\) = field without dielectric
- \(\mathrm{\kappa}\) = dielectric constant (\(\kappa \geq 1\))
Dielectric Constant and Permittivity:
The dielectric constant relates the permittivity of a material (\(\varepsilon\)) to the permittivity of free space (\(\varepsilon_0\)):
\( \mathrm{\kappa = \dfrac{\varepsilon}{\varepsilon_0}} \)
- \(\mathrm{\varepsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2)}\) (free-space permittivity).
- Higher \(\kappa\) means the material is more effective at reducing the field and increasing capacitance.
Capacitance with a Dielectric:
The capacitance of a parallel plate capacitor filled with a dielectric is:
\( \mathrm{C = \kappa C_0 = \kappa \dfrac{\varepsilon_0 A}{d} = \dfrac{\varepsilon A}{d}} \)
- \(\mathrm{C_0}\) = capacitance without dielectric
- \(\mathrm{A}\) = plate area
- \(\mathrm{d}\) = plate separation
Stored Charge and Energy:
If the capacitor is connected to a battery (\(V\) fixed):
- Capacitance increases → stored charge increases.
- Stored energy: \( \mathrm{U = \dfrac{1}{2} C V^2} \) increases.
If the capacitor is isolated (\(Q\) fixed):
- Capacitance increases → voltage decreases (\( V = Q/C \)).
- Stored energy decreases: \( \mathrm{U = \dfrac{Q^2}{2C}} \).
Key Features of Dielectrics:
- Reduce electric field between plates.
- Increase capacitance by factor \(\kappa\).
- Relate material permittivity to free-space permittivity: \( \mathrm{\kappa = \varepsilon / \varepsilon_0} \).
- Allow more energy storage when connected to a battery.
- Provide electrical insulation, preventing discharge.
Example:
A parallel plate capacitor has capacitance \( C_0 = 5.0 \, \mu F \). It is connected to a \( 12 \, V \) battery. A dielectric of \(\kappa = 3\) is inserted between the plates. Find the new charge stored and the new energy.
▶️ Answer/Explanation
Step 1: New capacitance: \( \mathrm{C = \kappa C_0 = 3 \times 5.0 = 15.0 \, \mu F} \)
Step 2: Stored charge: \( \mathrm{Q = C V = (15.0 \times 10^{-6})(12)} = 1.8 \times 10^{-4} \, C \)
Step 3: Stored energy: \( \mathrm{U = \tfrac{1}{2} C V^2 = 0.5 (15.0 \times 10^{-6})(144)} = 1.08 \times 10^{-3} \, J \)
Final Answer: \( Q = 1.8 \times 10^{-4} \, C \), \( U = 1.08 \times 10^{-3} \, J \).
Example:
A capacitor with \( C_0 = 4.0 \, \mu F \) is charged to \( Q = 8.0 \, \mu C \). It is then disconnected from the battery and filled with a dielectric of \(\kappa = 2\). Find the new potential difference and stored energy.
▶️ Answer/Explanation
Step 1: New capacitance: \( \mathrm{C = \kappa C_0 = 2 \times 4.0 = 8.0 \, \mu F} \)
Step 2: Voltage: \( \mathrm{V = \tfrac{Q}{C} = \tfrac{8.0 \times 10^{-6}}{8.0 \times 10^{-6}} = 1.0 \, V} \)
Step 3: Stored energy: \( \mathrm{U = \tfrac{Q^2}{2C} = \tfrac{(8.0 \times 10^{-6})^2}{2 (8.0 \times 10^{-6})}} = 4.0 \times 10^{-6} \, J \)
Final Answer: \( V = 1.0 \, V \), \( U = 4.0 \times 10^{-6} \, J \).