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Sagot :
Given
The fixed charge is q=62.92 mC
The other charge is q'=7.21 mC
The distance is r=8.57 cm=0.0857 m
To find
The work done
Explanation
The potential is
[tex]\begin{gathered} V=\frac{kq}{r} \\ \Rightarrow V=9\times10^9\frac{62.92\times10^{-6}}{0.0857} \\ \Rightarrow V=6.6\times10^6V \end{gathered}[/tex]Work done is
[tex]\begin{gathered} W=q^{\prime}V \\ \Rightarrow W=7.21\times10^{-6}\times6.6\times10^6=47.586\text{ J} \end{gathered}[/tex]Conclusion
The work done is 47.586 V
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