Finding the voltage when the electric charge and capacitance are known is fundamental in electronics. The relationship between these quantities can be determined using a specific formula and algebraic manipulation.
The Formula: \( C = \dfrac{Q}{V} \)
To find the voltage, we rearrange the formula as follows:
\[ V = \dfrac{Q}{C} \]
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: Voltage in a Camera Flash
Question: A camera flash has a capacitance of 200 microfarads (μF) and stores an electric charge of 0.04 coulombs. What is the operating voltage?
Calculation:
Given:
- \( C = 200 \times 10^{-6} \) F
- \( Q = 0.04 \) C
Using the formula:
\[ V = \dfrac{Q}{C} = \dfrac{0.04}{200 \times 10^{-6}} = 200 \, \text{V} \]
Result: The operating voltage of the camera flash is 200 volts.
Example 2: Voltage in a Defibrillator Capacitor
Question: A defibrillator capacitor has a capacitance of 100 microfarads (μF) and stores 0.05 coulombs of charge. What is the voltage across the capacitor?
Calculation:
Given:
- \( C = 100 \times 10^{-6} \) F
- \( Q = 0.05 \) C
Using the formula:
\[ V = \dfrac{Q}{C} = \dfrac{0.05}{100 \times 10^{-6}} = 500 \, \text{V} \]
Result: The voltage across the defibrillator capacitor is 500 volts.
Example 3: Voltage in a Tuning Circuit Capacitor
Question: A tuning circuit capacitor with a capacitance of 300 picofarads (pF) stores 3 nanocoulombs (nC) of charge. What is the voltage?
Calculation:
Given:
- \( C = 300 \times 10^{-12} \) F
- \( Q = 3 \times 10^{-9} \) C
Using the formula:
\[ V = \dfrac{Q}{C} = \dfrac{3 \times 10^{-9}}{300 \times 10^{-12}} = 10 \, \text{V} \]
Result: The voltage in the tuning circuit capacitor is 10 volts.
