Final answer:
The energy stored in a 7.00-mF capacitor with a charge of +/-50.0 mC is calculated using the formula E = (Q^2) / (2C), which results in 179 millijoules.
Step-by-step explanation:
The question asks to calculate the energy stored in a 7.00-mF (millifarad) air-filled capacitor with a charge of +/-50.0 mC (millicoulombs). To find the energy (E) stored in a capacitor, we can use the formula:
E = (Q^2) / (2C)
where Q is the charge stored on the capacitor and C is the capacitance of the capacitor.
Plugging in the given values:
E = (50.0 x 10^-3 C)^2 / (2 x 7.00 x 10^-3 F)
E = (2.50 x 10^-3 C^2) / (1.40 x 10^-2 F)
E = 0.179 J
E = 179 mJ
Therefore, the energy stored in the 7.00-mF capacitor is 179 millijoules (option A).