Question

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?

Answer 1

Step-by-step 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

2.4*10⁻⁵ J

Step-by-step 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