Answer:
the potential at a distance of 16 cm from the charge [tex]4\mu C[/tex] will be 225000 volt
Explanation:
We have given charge [tex]q=4\mu C=4\times 10^{-6}C[/tex]
Distance between the charge r = 16 cm = 0.16 m
Electric potential is given by [tex]V=\frac{1}{4\pi \varepsilon _0}\frac{q}{r}=\frac{Kq}{r}[/tex], here K is constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]
So potential [tex]V=\frac{9\times 10^9\times 4\times 10^{-6}}{0.16}=225000volt[/tex]
So the potential at a distance of 16 cm from the charge [tex]4\mu C[/tex] will be 225000 volt
