Sure! Let's solve this step by step.
1. Understanding the given values:
- The potential difference (V) between the two wires is \(120\, \text{V}\).
- The charge (Q) on the wires is \(7.5 \times 10^{-10}\, \text{C}\), wherein they hold equal and opposite charges.
2. Applying the formula for capacitance:
- The capacitance (C) of a system can be calculated using the formula:
[tex]\[
C = \frac{Q}{V}
\][/tex]
- Here, \(Q\) is the charge and \(V\) is the potential difference.
3. Substitute the given values into the formula:
- \(Q = 7.5 \times 10^{-10} \text{ C}\)
- \(V = 120 \text{ V}\)
Substituting these:
[tex]\[
C = \frac{7.5 \times 10^{-10} \text{ C}}{120 \text{ V}} = 6.25 \times 10^{-12} \text{ F}
\][/tex]
4. Comparing with the multiple-choice options:
- Now, we compare the calculated capacitance \(6.25 \times 10^{-12} \text{ F}\) with the given options:
- \(3.25 \times 10^{-7} \text{ F}\)
- \(3.4 \times 10^{-26} \text{ F}\)
- \(4.25 \times 10^{-6} \text{ F}\)
- \(6.25 \times 10^{-12} \text{ F}\)
- The correct match is \(6.25 \times 10^{-12} \text{ F}\).
Therefore, the capacitance of the two wires is [tex]\(\boxed{6.25 \times 10^{-12} \text{ F}}\)[/tex].
