A point of charge 10×10^-6C has a potential of 4×10^4V from a point 2mm from it. Calculate the potential due to the charge when the distance is now 16mm.
Given:
"V_1=4*10^4\\:\\rm V"
"r_1=2*10^{-3}\\:\\rm m"
"r_2=16*10^{-3}\\:\\rm m"
The potential of electric field of point charge
"V=k\\frac{q}{r}\\sim\\frac{1}{r}"Hence
"V_2=V_1\\frac{r_1}{r_2}=4*10^4*\\frac{0.002}{0.016}=5000\\:\\rm V"
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