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?

Question

Phelps

Physics

20 September, 02:58

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?

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Answers (2)

Know the Answer?

Ben Juarez · Answer 1

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

Jessica Davies · Answer 2

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