When a 12.0-V battery causes 2.00 μC of charge to flow onto the plates of an air-filled capacitor, how much work did the battery
do?
2 answers:
Answer:
Explanation:
Voltage, V = 12 V
Charge, q = 2 micro coulomb = 2 x 10^-6 C
Work = energy
W = 0.5 x q x V
W = 0.5 x 2 x 10^-6 x 12
W = 12 x 10^-6 J
Answer:
2.4*10⁻⁵ J
Explanation:
workdone in moving a charge across a potential difference = charge * the potential difference across.
w = qV
q = 2.0μC = 2.0*10⁻⁶C
V = 12.0V
W = 2.0*10⁻⁶ * 12.0
W = 2.4*10⁻⁵ J
The battery did a work of 2.4*10⁻⁵ J
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