To find the electrical potential energy stored in the capacitor, we can use the formula:
[tex]\[ PE_e = \frac{1}{2} Q V \][/tex]
where:
- [tex]\(PE_e\)[/tex] is the electrical potential energy,
- [tex]\(Q\)[/tex] is the charge on each plate,
- [tex]\(V\)[/tex] is the potential difference across the plates.
Given values:
- [tex]\( Q = 8.0 \times 10^{-10} \, \text{C} \)[/tex]
- [tex]\( V = 40.0 \, \text{V} \)[/tex]
Plug these values into the formula:
[tex]\[ PE_e = \frac{1}{2} \times (8.0 \times 10^{-10} \, \text{C}) \times (40.0 \, \text{V}) \][/tex]
First, perform the multiplication within the parentheses:
[tex]\[ 8.0 \times 10^{-10} \, \text{C} \times 40.0 \, \text{V} = 320.0 \times 10^{-10} \, \text{CV} \][/tex]
Then, multiply by [tex]\( \frac{1}{2} \)[/tex]:
[tex]\[ PE_e = \frac{1}{2} \times 320.0 \times 10^{-10} \, \text{CV} \][/tex]
[tex]\[ PE_e = 160.0 \times 10^{-10} \, \text{J} \][/tex]
Convert [tex]\( 160.0 \times 10^{-10} \, \text{J} \)[/tex] to a more standard scientific notation:
[tex]\[ 160.0 \times 10^{-10} \, \text{J} = 1.6 \times 10^{-8} \, \text{J} \][/tex]
Thus, the electrical potential energy stored in the capacitor is:
[tex]\[ \boxed{1.6 \times 10^{-8} \, \text{J}} \][/tex]
Hence, the correct option is:
A. [tex]\(1.6 \times 10^{-8} \, \text{J}\)[/tex]
