? asked in Science & MathematicsPhysics · 1 month ago

what work is done when 0.024 c of charge is raised in potential by 3.3 v?

5 Answers

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  • MyRank
    Lv 6
    1 month ago

    Given,

    Charge (q) = 0.024 C

    Potential difference (∆v) = 3.3v

    Work done (w) = ?

    We know that:-

    Work done (w) = charge (q) x potential difference (∆v)

    = 0.024c x 3.3 v

    = 0.0792J.

  • 1 month ago

    delta(E) = W = Q delta(v) = 0.024 C * 3.3 J/C = 0.0792 J

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  • Anonymous
    1 month ago

    Work = 0.024/2*3.3^2 = 0.13 joule

  • 1 month ago

    Do you mean 0.024 C ?

    E = ½QV = ½0.024²3.3 = 0.00095 JEnergy in a Capacitor in JoulesE = ½CV² = ½QV = ½Q²/CQ = CVQ is charge in coulombsC is capacitance in FaradsV is voltage in voltsE is energy in Joules

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  • Mangal
    Lv 4
    1 month ago

    Work done = q∆V = 0.024 x 3.3 = 0.0792 Joule (ans)

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