Let the charge is q,
so electric potential at the distance r is
V=q4πϵorV=\frac{q}{4\pi \epsilon_o r}V=4πϵorq
if r is infinite then potential (V)=q4π∞=0(V)=\frac{q}{4\pi \infty}=0(V)=4π∞q=0
Now, if(r)=1.00mm=1.0×10−3m( r)=1.00mm=1.0\times 10^{-3}m(r)=1.00mm=1.0×10−3m
V=q4πϵo×1.0×10−3mV=\frac{q}{4\pi \epsilon_o\times 1.0\times 10^{-3}m}V=4πϵo×1.0×10−3mq
=q×1034πϵo=\frac{q\times 10^{3}}{4\pi \epsilon_o}=4πϵoq×103
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