Final answer:
The voltage required to store 0.160 mC of charge on an 8.00 nF capacitor is 20.0 V, which is option (a). This is determined by using the formula V = Q/C and ensuring that the units for charge and capacitance are consistent before performing the calculation.
Step-by-step explanation:
To determine the voltage required to store 0.160 mC of charge on an 8.00 nF capacitor, we can use the fundamental relationship between charge (Q), capacitance (C), and voltage (V), which is expressed by the formula V = \( \frac{Q}{C} \). First, we'll need to ensure the units are consistent, converting 0.160 mC (milliCoulombs) to Coulombs by multiplying by 10-3, and converting 8.00 nF (nanofarads) to farads by multiplying by 10-9. Substituting the converted values into the formula, we get:
- Q = 0.160 × 10-3 C = 1.60 × 10-4 C
- C = 8.00 × 10-9 F
Plugging the values into the formula:
V = \( \frac{1.60 \times 10^{-4} C}{8.00 \times 10^{-9} F} \) = \( \frac{1.60 \times 10^{-4}}{8.00 \times 10^{-9}} \)
Calculating this gives us:
V = 20.0 V
Thus, the voltage that must be applied to an 8.00 nF capacitor to store 0.160 mC of charge is (20 V), which corresponds to option (a).