Determining the capacitance when the electric charge and voltage are known is crucial in various electronic devices. The relationship between these quantities can be calculated using a specific formula and algebraic manipulation.
The Formula: \( C = \dfrac{Q}{V} \)
Where:
- \( C \) is the capacitance (measured in farads, F)
- \( Q \) is the electric charge (measured in coulombs, C)
- \( V \) is the voltage (measured in volts, V)
Example 1: Capacitance of a Smartphone Battery
Question: A smartphone battery stores an electric charge of 3000 milliampere-hours (mAh), which is approximately 10,800 coulombs, and operates at 3.7 volts. What is the battery's capacitance?
Calculation:
Given:
- \( Q = 10,800 \) C
- \( V = 3.7 \) V
Using the formula:
\[ C = \dfrac{Q}{V} = \dfrac{10,800}{3.7} \approx 2918.92 \, \text{F} \]
Result: The capacitance of the smartphone battery is approximately 2918.92 farads.
Example 2: Capacitance of a Flashlight Capacitor
Question: A flashlight capacitor stores 0.002 coulombs of charge at a voltage of 5 volts. What is the capacitance?
Calculation:
Given:
- \( Q = 0.002 \) C
- \( V = 5 \) V
Using the formula:
\[ C = \dfrac{Q}{V} = \dfrac{0.002}{5} = 0.0004 \, \text{F} = 400 \, \text{μF} \]
Result: The capacitance of the flashlight capacitor is 400 microfarads.
Example 3: Capacitance of a Tuning Circuit Capacitor
Question: A tuning circuit capacitor stores 1 nanocoulomb (nC) of charge at a voltage of 2 volts. What is the capacitance?
Calculation:
Given:
- \( Q = 1 \times 10^{-9} \) C
- \( V = 2 \) V
Using the formula:
\[ C = \dfrac{Q}{V} = \dfrac{1 \times 10^{-9}}{2} = 0.5 \times 10^{-9} \, \text{F} = 0.5 \, \text{nF} \]
Result: The capacitance of the tuning circuit capacitor is 0.5 nanofarads.
