AP Physics 2- 10.6 Capacitors- Study Notes- New Syllabus
AP Physics 2- 10.6 Capacitors – Study Notes
AP Physics 2- 10.6 Capacitors – Study Notes – per latest Syllabus.
Key Concepts:
- Parallel-Plate Capacitor
- Electric Field Between Parallel Plates
- Electric Potential Energy Stored in a Capacitor
Parallel-Plate Capacitor
A parallel-plate capacitor consists of two parallel conducting surfaces separated by a small distance. These plates can hold equal amounts of charge with opposite signs. The capacitor stores energy in the electric field created between the plates.
Capacitance:
Capacitance (\(C\)) is the measure of a capacitor’s ability to store charge per unit potential difference:
\( C = \dfrac{Q}{V} \)
where \(Q\) is the magnitude of charge on each plate and \(V\) is the potential difference between the plates.
The capacitance depends only on the physical properties of the capacitor, not on the charge or voltage applied.
Physical Factors Affecting Capacitance:
Plate Area (A):
- Capacitance is directly proportional to the area of one plate: larger area → more charge storage.
Separation Between Plates (d):
- Capacitance is inversely proportional to the distance between plates: smaller separation → stronger field → higher capacitance.
Dielectric Material:
- The dielectric constant (\( \varepsilon_r \)) of the material between the plates increases the capacitance.
- The product of dielectric constant and permittivity of free space gives the proportionality constant:
\( C = \varepsilon_0 \varepsilon_r \dfrac{A}{d} \)
Example :
A parallel-plate capacitor has plate area \( 0.01 \, m^2 \), separation \( 2 \, mm \), and air between the plates. Find its capacitance.
▶️ Answer/Explanation
Formula: \( C = \varepsilon_0 \dfrac{A}{d} \)
= \( (8.85 \times 10^{-12}) \dfrac{0.01}{0.002} \)
= \( 4.425 \times 10^{-11} \, F \)
Answer: Capacitance ≈ \( 44.25 \, pF \).
Electric Field Between Parallel Plates
For two charged parallel plates with uniformly distributed charge, the electric field is constant in both magnitude and direction between the plates, except near the edges (fringing effects). This approximation is valid when the plate separation is much smaller than the dimensions of the plates.
Electric Field Magnitude:
For plates with potential difference \(V\) and separation \(d\):
\( E = \dfrac{V}{d} \)
- Direction: From positive plate to negative plate.
- Uniform field → field lines are parallel and equally spaced between the plates.
The magnitude of the electric field between two charged parallel plates, where the plate separation is much smaller than the dimensions of the plates, can also be described as:
\( E_C = \dfrac{Q}{\kappa \varepsilon_0 A} \)
where \(Q\) is the charge on a plate, \(A\) is the plate area, \(\kappa\) is the dielectric constant (for vacuum or air, \(\kappa = 1\)), and \(\varepsilon_0\) is the permittivity of free space.
Motion of a Charged Particle Between Plates:
A charged particle placed in the uniform electric field experiences a constant force:
\( F = qE \)
Since \( F = ma \), the particle undergoes constant acceleration along the direction of the field:
\( a = \dfrac{qE}{m} \)
The motion of the particle shares characteristics with projectile motion in a uniform gravitational field near Earth:
- Horizontal motion (if any) remains uniform.
- Vertical motion (along the field) is accelerated.
Example :
A proton is released from rest between two parallel plates separated by 0.01 m, with a potential difference of 500 V. Find the acceleration of the proton.
▶️ Answer/Explanation
Step 1: Electric field between plates: \( E = \dfrac{V}{d} = \dfrac{500}{0.01} = 5.0 \times 10^4 \, N/C \)
Step 2: Force on proton: \( F = qE = (1.6 \times 10^{-19})(5.0 \times 10^4) = 8.0 \times 10^{-15} \, N \)
Step 3: Acceleration: \( a = \dfrac{F}{m} = \dfrac{8.0 \times 10^{-15}}{1.67 \times 10^{-27}} \approx 4.79 \times 10^{12} \, m/s^2 \)
Answer: Acceleration ≈ \( 4.8 \times 10^{12} \, m/s^2 \).
Electric Potential Energy Stored in a Capacitor
The electric potential energy (\(U\)) stored in a capacitor is equal to the work done by an external force to separate charge on the plates against the electric field. This energy is stored in the electric field between the plates.
Key Equations:
Energy stored in a capacitor:
\( U = \dfrac{1}{2} C V^2 \)
where \(C\) is capacitance and \(V\) is potential difference.
Alternative forms using charge (\(Q\)) and capacitance:
\( U = \dfrac{Q^2}{2C} \)
\( U = \dfrac{1}{2} Q V \)
Effect of a Dielectric:
Inserting a dielectric between the plates:
- Increases the capacitance: \( C = \varepsilon_0 \varepsilon_r \dfrac{A}{d} \)
- Reduces the effective electric field inside the capacitor:
\( E_{effective} = \dfrac{E_{plate}}{\varepsilon_r} \)
The dielectric becomes polarized, creating an induced electric field in the opposite direction to the field between the plates. With a dielectric, more charge can be stored at the same potential difference, increasing the stored energy.
Example :
A 10 μF capacitor is charged to 12 V. Find the energy stored in the capacitor.
▶️ Answer/Explanation
Formula: \( U = \dfrac{1}{2} C V^2 \)
= \( 0.5 \times 10 \times 10^{-6} \times 12^2 \)
= \( 7.2 \times 10^{-4} \, J \)
Answer: Energy stored = 0.72 mJ.