What is the capacitance of a capacitor that stores 5.2 μC of charge on its plates when a voltage of 4.8 V is applied between them?
To find the capacitance of a capacitor, we can use the formula:
C = Q/V
Where:
- C is the capacitance of the capacitor
- Q is the charge stored on the plates of the capacitor
- V is the voltage applied across the plates of the capacitor
In this case, we are given that the charge stored on the plates is 5.2 μC (microcoulombs) and the voltage applied is 4.8 V.
Now let's substitute these values into the formula to find the capacitance:
C = 5.2 μC / 4.8 V
Since the charge is given in microcoulombs, we need to convert it to coulombs first. There are 1 million microcoulombs (μC) in 1 coulomb (C). So, 5.2 μC is equal to 5.2 * 10^(-6) C.
C = (5.2 * 10^(-6) C) / 4.8 V
Simplifying this expression, we can calculate:
C ≈ 1.0833 * 10^(-6) C / V
Therefore, the capacitance of the capacitor is approximately 1.0833 microfarads (μF).