Ask Question
24 March, 20:05

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?

+5
Answers (2)
  1. 24 March, 20:21
    0
    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
  2. 24 March, 23:41
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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? ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers