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19 December, 15:55

What is the emf of a battery that increases the electric potential energy of 0.060 CC of charge by 0.70 JJ as it moves it from the negative to the positive terminal?

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Answers (2)
  1. 19 December, 18:31
    0
    11.67 V.

    Explanation:

    Workdone by a charge, W = q * v

    Where,

    W = 0.7 J

    q = 0.06 C

    v = W/q

    = 0.7/0.06

    = 11.67 V

    Emf = 11.67 V
  2. 19 December, 19:32
    0
    emf = 11.667 V

    Explanation:

    Given: charge q = 0.060 C, electric potential energy E = 0.70 J,

    Solution:

    by definition 1 volt = 1 joule per coulomb

    so Voltage = emf = E/C

    emf = 0.70 J / 0.060 C

    emf = 11.667 V
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