A charge of 6 × 10−7𝐶 is transferred from infinity to point B. If the work
done by the electric force to do this is 1.2 × 10−5𝐽. What is the potential at point
B?
Answer
the potential at point
V=Wq=1.2∗10−56∗10−7=20VoltsV=\frac{W}{q}\\=\frac{1.2*10^{-5}}{6*10^{-7}}\\=20VoltsV=qW=6∗10−71.2∗10−5=20Volts
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