Calculating the electric charge when the capacitance and voltage are known is essential in many electronic applications. The relationship between these quantities can be expressed using a specific formula and algebraic manipulation.
The Formula: \( C = \dfrac{Q}{V} \)
To find the electric charge, we rearrange the formula as follows:
\[ Q = C \cdot V \]
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: Charge in a Camera Flash
Question: A camera flash has a capacitance of 100 microfarads (μF) and is charged to 300 volts. What is the stored electric charge?
Calculation:
Given:
- \( C = 100 \times 10^{-6} \) F
- \( V = 300 \) V
Using the formula:
\[ Q = C \cdot V = 100 \times 10^{-6} \cdot 300 = 0.03 \, \text{C} \]
Result: The stored electric charge in the camera flash is 0.03 coulombs.
Example 2: Charge in a Capacitor of a Defibrillator
Question: A defibrillator uses a capacitor with a capacitance of 50 microfarads (μF) and a voltage of 2000 volts. What is the electric charge stored in the capacitor?
Calculation:
Given:
- \( C = 50 \times 10^{-6} \) F
- \( V = 2000 \) V
Using the formula:
\[ Q = C \cdot V = 50 \times 10^{-6} \cdot 2000 = 0.1 \, \text{C} \]
Result: The electric charge stored in the defibrillator's capacitor is 0.1 coulombs.
Example 3: Charge in a Tuning Circuit
Question: A tuning circuit in a radio has a capacitance of 500 picofarads (pF) and a voltage of 10 volts. What is the electric charge?
Calculation:
Given:
- \( C = 500 \times 10^{-12} \) F
- \( V = 10 \) V
Using the formula:
\[ Q = C \cdot V = 500 \times 10^{-12} \cdot 10 = 5 \times 10^{-9} \, \text{C} = 5 \, \text{nC} \]
Result: The electric charge in the tuning circuit is 5 nanocoulombs.
