Respuesta :
Given that the radius of the conducting sphere is r = 5 cm = 0.05 m
The charge density on the surface of the sphere is
[tex]\sigma\text{ = 2}\times\frac{10^{-6}C}{m^2}[/tex]We have to find the electric potential.
To calculate electric potential, first, we need to calculate charge.
The charge will be
[tex]\begin{gathered} \sigma=\frac{Q}{4\pi r^2} \\ Q=4\pi r^2\times\sigma \\ =\text{ 4}\times3.14\times(0.05)^2\times2\times10^{-6} \\ =\text{ 6.28}\times10^{-8}\text{ C} \end{gathered}[/tex]The electric potential is given by the formula
[tex]V=\text{ k}\frac{Q}{r}[/tex]Here K is Coulomb's constant whose value is
[tex]K=\text{ 8.99 }\times\frac{10^9Nm^2}{C^2}[/tex]Substituting the values, the electric potential will be
[tex]\begin{gathered} V=\text{ }\frac{8.99\times10^9\times6.28\times10^{-8}}{0.05} \\ =11291.44 \\ =1.1\times10^4\text{ V} \end{gathered}[/tex]
