[tex]4.8 \times 10^8[/tex] volts
Explanation:
The electric potential due to a point charge is given by
[tex]V= \dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q}{r}[/tex]
where Q = charge = [tex]16 \times 10^{-9}[/tex] C
r = distance from a point = [tex]3 \times 10^{-3}[/tex] m
[tex]\varepsilon_{0}[/tex] = permitivity of free space
= 8.85×10^-12 C^2/N-m^2
Plugging in the numbers,
[tex]V = \dfrac{1}{4 \pi (8.85 \times 10^{-12})} \dfrac{16 \times 10{-9}}{3 \times 10^{-3}}[/tex]
[tex]= 4.8 \times 10^8[/tex] volts
