When a 12.0-V battery causes $2.00 \mu \mathrm { C }$ of charge to flow onto the plates of an air-filled capacitor, how much work did the battery do?

Question

Quizplus · Accepted Answer

The work done by the battery is given by $ W = QV $, where $ Q = 2.00 \mu \mathrm{C} $ and $ V = 12.0 \mathrm{V} $. Thus, $ W = 2.00 \times 10^{-6} \mathrm{C} \times 12.0 \mathrm{V} = 24.0 \times 10^{-6} \mathrm{J} = 24.0 \mu \mathrm{J} $.

When a 120-V Battery Causes 2.00μC2.00 \mu \mathrm { C }2.00μC

When a 120-V Battery Causes $2.00 \mu \mathrm { C }$