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How much work is needed to transfer 767.1 kC of charge in one hour through a potential rise of 567.78 V ?

User Endbegin
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1 Answer

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The change in potential energy ΔU of a system where a charge q moves through a potential difference ΔV is:


\Delta U=q\Delta V

The work needed to move a charge q through a potential difference ΔV is the same as the change in potential energy of the system. Then, replace q=767.1kC and ΔV=567.78V to find the work needed to complete the described process:


\begin{gathered} \Delta U=q\Delta V \\ =(767.1kC)(567.78V) \\ =435,544kJ \end{gathered}

Therefore, the amount of work needed to complete such a process is approximaely 435,544 kiloJoules.

User Rahmu
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