Question

The two plates of a capacitor hold +9.5x10^-6 C and -9.5 x10^-6 C of charge when the potential difference is 725 V. What is the capacitance?
Group of answer choices

3.4x10^-26 F

2.6x10^-7 F

1.3x10^-8 F

9.9x10^-6 F

Answer 1

The capacitance \( C \) of a capacitor can be calculated using the formula:

\[
C = \frac{Q}{V}
\]

where:
- \( Q \) is the charge (in coulombs),
- \( V \) is the potential difference (in volts).

In this case, the charge \( Q \) is given as \( 9.5 \times 10^{-6} \) C, and the potential difference \( V \) is \( 725 \) V.

Now we can substitute the values into the formula:

\[
C = \frac{9.5 \times 10^{-6} \, \text{C}}{725 \, \text{V}}
\]

Calculating this gives:

\[
C = \frac{9.5 \times 10^{-6}}{725} \approx 1.31034 \times 10^{-8} \, \text{F}
\]

Rounding this value, we find:

\[
C \approx 1.3 \times 10^{-8} \, \text{F}
\]

So the correct answer from the given choices is:

**1.3 x 10^-8 F**