Section 9.3: Capacitance
Capacitance is the ability of a system to store electric charge per unit potential difference. It is a measure of how much charge a conductor can hold for a given voltage.
\[
C = \frac{Q}{V}
\]
where:
\( C \) = capacitance (F, farads)
\( Q \) = charge stored (C)
\( V \) = potential difference across the plates (V)
\( C \) = capacitance (F, farads)
\( Q \) = charge stored (C)
\( V \) = potential difference across the plates (V)
For a parallel plate capacitor, the capacitance is given by:
\[
C = \varepsilon_0 \frac{A}{d}
\]
where:
\( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) (permittivity of free space)
\( A \) = plate area (m²)
\( d \) = separation between plates (m)
\( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) (permittivity of free space)
\( A \) = plate area (m²)
\( d \) = separation between plates (m)
Example 1: Charge on a Capacitor
A 6 μF capacitor is connected to a 12 V battery. Find the charge stored.
\( Q = C V = 6 \times 10^{-6} \cdot 12 = 72 \times 10^{-6} \, \text{C} = 72 \, \mu\text{C} \)
Example 2: Capacitance of Parallel Plates
A parallel plate capacitor has area 0.02 m² and separation 0.001 m. Calculate its capacitance.
\( C = \varepsilon_0 \frac{A}{d} = 8.85\times10^{-12} \frac{0.02}{0.001} \approx 1.77 \times 10^{-10} \, \text{F} \)
Practice Problems
- A 10 μF capacitor is charged to 5 V. Find the charge stored.
- A parallel plate capacitor has capacitance 4 μF, plate area 0.01 m². Find the separation if the space is vacuum.
- Two capacitors of 6 μF and 12 μF are connected in series. Find the equivalent capacitance.
- A 3 μF capacitor is charged to 9 V. Calculate the energy stored in the capacitor.
- Calculate the capacitance of a parallel plate capacitor with plate separation 0.002 m and plate area 0.05 m² in air.