In moving a charge of 3 C across a battery, 300 J of energy is dissipated. If the negative terminal of the battery is at an electric potential of –20 V, then the electric potential of the positive terminal isOnly one correct answerA.40 VB.20 VC.60 VD.80 V
Question
In moving a charge of 3 C across a battery, 300 J of energy is dissipated. If the negative terminal of the battery is at an electric potential of –20 V, then the electric potential of the positive terminal isOnly one correct answerA.40 VB.20 VC.60 VD.80 V
Solution
The energy dissipated (W) when moving a charge (Q) across a battery is given by the equation:
W = Q * V
where V is the potential difference across the battery. We can rearrange this equation to solve for V:
V = W / Q
Substituting the given values:
V = 300 J / 3 C = 100 V
This is the potential difference across the battery. Since the negative terminal is at -20 V, the positive terminal must be at a higher potential to give a difference of 100 V. Therefore, the potential at the positive terminal is:
-20 V + 100 V = 80 V
So, the correct answer is D. 80 V.
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