A parallel-plate capacitor with plate area 2.0 cm² and air-gap separation 0.50 mm is connected to a 12-V battery, and fully charged. The battery is then disconnected. What is the charge on the capacitor?

Verified step by step guidance

1. Convert the given values into SI units: The plate area is 2.0 cm², which is equivalent to 2.0 × 10⁻⁴ m². The air-gap separation is 0.50 mm, which is equivalent to 0.50 × 10⁻³ m.
2. Recall the formula for the capacitance of a parallel-plate capacitor: $C=\frac{\epsilon }{d}\cdot A$, where $\epsilon$ is the permittivity of free space (8.85 × 10⁻¹² F/m), $A$ is the plate area, and $d$ is the separation between the plates.
3. Substitute the values into the capacitance formula: $C=\frac{8.85\times {10}^{-}}{0.50\times {10}^{-}}\cdot 2.0\times {10}^{-}$.
4. Once the capacitance is determined, use the formula for the charge on a capacitor: $Q=C\cdot V$, where $V$ is the voltage across the capacitor (12 V in this case).
5. Substitute the calculated capacitance and the given voltage into the charge formula to find the charge on the capacitor.

Key Concepts

Capacitance

Capacitance is the ability of a capacitor to store charge per unit voltage. It is defined as C = Q/V, where C is capacitance, Q is the charge stored, and V is the voltage across the capacitor. For a parallel-plate capacitor, capacitance can also be calculated using the formula C = ε₀(A/d), where ε₀ is the permittivity of free space, A is the plate area, and d is the separation between the plates.

Charge on a Capacitor

The charge on a capacitor is the amount of electric charge stored on its plates when a voltage is applied. Once the capacitor is charged and the battery is disconnected, the charge remains constant. The relationship between charge, capacitance, and voltage allows us to calculate the charge using the formula Q = C × V, where Q is the charge, C is the capacitance, and V is the voltage.

Units of Measurement

Understanding the units of measurement is crucial in physics. Capacitance is measured in farads (F), charge in coulombs (C), and voltage in volts (V). In this problem, the area is given in square centimeters (cm²) and the separation in millimeters (mm), which may require unit conversions to ensure consistency when calculating capacitance and charge.