Charging a Parallel-Plate Capacitor

Problem Statement

Two square plates of area 200 cm^2 are separated by 2 mm of air. The plates are connected to a 120 V battery.

1. Find the capacitance of the setup.
2. Determine the charge stored on each plate once the capacitor is fully charged.
3. Compute the energy stored in the electric field.

(Take epsilon_0 = 8.85 x 10^-12 F/m.)

Given / Required

• A = 200 cm^2 = 200 x 10^-4 m^2 = 0.02 m^2.
• d = 2 mm = 2 x 10^-3 m.
• V = 120 V.

Need to find C, Q and U.

Hint

Treat the plates as an ideal parallel-plate capacitor: \(C = \varepsilon_0 A / d\). Stored charge equals CV and energy is 0.5 C V^2.

Step-by-Step Solution

Step 1 – Capacitance

\[ C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.02}{2 \times 10^{-3}} \approx 8.85 \times 10^{-11},\text{F} \]

Step 2 – Charge on Each Plate

\[ Q = C V = 8.85 \times 10^{-11} \times 120 \approx 1.06 \times 10^{-8},\text{C} \]

Each plate stores this amount with opposite signs.

Step 3 – Stored Energy

\[ U = \tfrac{1}{2} C V^2 = 0.5 \times 8.85 \times 10^{-11} \times 120^2 \approx 6.37 \times 10^{-7},\text{J} \]

Final Answer

• Capacitance: roughly 8.9 x 10^-11 F.
• Charge magnitude: about 1.1 x 10^-8 C on each plate.
• Energy stored: about 6.4 x 10^-7 J.

Extension / Variation

• Replace air with a dielectric of relative permittivity 4.
• Decrease plate separation while keeping voltage fixed and note how energy changes.

Key Concept Recap

• Parallel-plate capacitance grows with area and shrinks with separation.
• Charge equals capacitance times applied voltage.
• Energy stored scales with the square of voltage.