what work is done when 0.024 c of charge is raised in potential by 3.3 v?
- MyRankLv 61 month ago
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.Source(s): https://myrank.co.in/
- Old Science GuyLv 71 month ago
delta(E) = W = Q delta(v) = 0.024 C * 3.3 J/C = 0.0792 J
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- Anonymous1 month ago
Work = 0.024/2*3.3^2 = 0.13 joule
- billrussell42Lv 71 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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- MangalLv 41 month ago
Work done = q∆V = 0.024 x 3.3 = 0.0792 Joule (ans)