To determine the voltage required to store a given charge on a capacitor, we use the relationship between charge (Q), capacitance (C), and voltage (V), which is given by:
[tex]\[ V = \frac{Q}{C} \][/tex]
Where:
- [tex]\( Q \)[/tex] is the charge in coulombs (C)
- [tex]\( C \)[/tex] is the capacitance in farads (F)
- [tex]\( V \)[/tex] is the voltage in volts (V)
Given:
- [tex]\( Q = 7.2 \times 10^{-5} \)[/tex] coulombs
- [tex]\( C = 6.0 \)[/tex] microfarads ([tex]\( \mu F \)[/tex])
First, we need to convert the capacitance from microfarads to farads since the standard unit for capacitance is farads. We know that [tex]\( 1 \mu F = 10^{-6} F \)[/tex], hence:
[tex]\[ C = 6.0 \mu F = 6.0 \times 10^{-6} \text{ F} \][/tex]
Now, substituting the given values into the formula:
[tex]\[ V = \frac{7.2 \times 10^{-5}}{6.0 \times 10^{-6}} \][/tex]
By performing the division:
[tex]\[ V = \frac{7.2 \times 10^{-5}}{6.0 \times 10^{-6}} = \frac{7.2}{6.0} \times \frac{10^{-5}}{10^{-6}} \][/tex]
[tex]\[ V = 1.2 \times 10^1 \][/tex]
[tex]\[ V = 12 \text{ V} \][/tex]
Therefore, the voltage required to store [tex]\(7.2 \times 10^{-5} \text{ C}\)[/tex] of charge on the plates of a [tex]\( 6.0 \mu F \)[/tex] capacitor is [tex]\( 12 \text{ V} \)[/tex].
Hence, the correct answer is:
A) 12 V
