Understanding how to find the electric charge using electric capacity and voltage is crucial for anyone working with electrical systems, from engineers to hobbyists. The relationship between electric capacity (C), electric charge (Q), and voltage (V) is given by the formula:
\[ C = \dfrac{Q}{V} \]
Where:
- \( C \) is the electric capacity (measured in farads, F)
- \( Q \) is the electric charge (measured in coulombs, C)
- \( V \) is the voltage (measured in volts, V)
To find the electric charge, we rearrange the formula:
\[ Q = C \cdot V \]
Example 1: Capacitor in a Flashlight
Question: A capacitor in a flashlight has a capacity of 0.01 farads and is charged to a voltage of 9 volts. How much electric charge does it store?
Calculation:
Given:
- \( C = 0.01 \) F
- \( V = 9 \) V
Using the formula:
\[ Q = 0.01 \cdot 9 = 0.09 \text{ coulombs} \]
Result: The capacitor in the flashlight stores 0.09 coulombs of electric charge.
Example 2: Electric Car Battery Pack
Question: An electric car battery pack has a total capacity of 85 farads and operates at 400 volts. What is the total electric charge it can hold?
Calculation:
Given:
- \( C = 85 \) F
- \( V = 400 \) V
Using the formula:
\[ Q = 85 \cdot 400 = 34,000 \text{ coulombs} \]
Result: The electric car battery pack can hold 34,000 coulombs of electric charge.
Example 3: Smartphone Battery
Question: A smartphone battery has a capacity of 0.0025 farads and is charged to 3.7 volts. How much electric charge does the battery store?
Calculation:
Given:
- \( C = 0.0025 \) F
- \( V = 3.7 \) V
Using the formula:
\[ Q = 0.0025 \cdot 3.7 = 0.00925 \text{ coulombs} \]
Result: The smartphone battery stores 0.00925 coulombs of electric charge.
These examples illustrate how to apply the formula to calculate the electric charge in different real-life scenarios. Whether you're working with capacitors in flashlights, battery packs in electric cars, or batteries in smartphones, understanding this relationship is essential for effective electrical system design and analysis.
