Find the distance of an isolated positive point charge of 10 nC, in order for it to produce
an electric potential of 120 V.
Answer
the distance of an isolated positive point charge of 10 nC, in order for it to produce
r=KqV=9∗109∗10∗10−9120=0.75m=75cmr=\frac{Kq}{V}\\=\frac{9*10^9*10*10^{-9}}{120}\\=0.75m\\=75cmr=VKq=1209∗109∗10∗10−9=0.75m=75cm
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