How much work is required to carry a 6 micro coulomb charge from negative to positive terminal of a 9 V battery ?

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Solution

Dear Student,

$charge, q = 6 μC = 6 \times 10^{- 6} C potential difference, V = 9 V Work done, W = qV W = 6 \times 10^{- 6} \times 9 = 54 \times 10^{- 6} J$

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How much work is required to carry a 6 micro coulomb charge from negative to positive terminal of a 9 V battery ?

Dear Student, charge, q=6 μC=6×10-6 Cpotential difference, V=9 VWork done, W=qVW=6×10-6×9=54×10-6 J

Dear Student,

$charge, q = 6 μC = 6 \times 10^{- 6} C potential difference, V = 9 V Work done, W = qV W = 6 \times 10^{- 6} \times 9 = 54 \times 10^{- 6} J$