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Question
In an ideal conductor without collisions, which relation connects the change in kinetic energy to the change in potential energy?
Options
\[\Delta K = \Delta U_{\text{pot}} = -IV\Delta t\]
\[\Delta K = \Delta U_{\text{pot}} = IV\Delta t\]
\[\Delta K = -\Delta U_{\text{pot}} = IV\Delta t\]
\[\Delta K = -\Delta U_{\text{pot}} = \frac{IV}{\Delta t}\]
MCQ
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Solution
With no collisions, the increase in kinetic energy equals the decrease in potential energy. Thus, \[\Delta K = -\Delta U_{\text{pot}} = IV\Delta t\].
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