Respuesta :
Answer:
(a). The electric potential at 1.650 cm is [tex]-1.219\times10^{-4}\ V[/tex].
(b). The electric potential at 2.81 cm is [tex]-3.759\times10^{-4}\ V[/tex].
Explanation:
Given that,
Radius of sphere R=2.81 cm
Charge = +2.35 fC
Potential at center of sphere
[tex]V = 0[/tex]
(a). We need to calculate the potential at a distance r = 1.60 cm
Using formula of potential difference
[tex]V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr[/tex]
[tex]V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr[/tex]
[tex]V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}[/tex]
[tex]V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})[/tex]
[tex]V_{r}=-0.00012190\ V[/tex]
[tex]V_{r}=-1.219\times10^{-4}\ V[/tex]
The electric potential at 1.650 cm is [tex]-1.219\times10^{-4}\ V[/tex].
(b). We need to calculate the potential at a distance r = R
Using formula of  potential difference
[tex]V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}[/tex]
[tex]V_{R}=-0.0003759\ V[/tex]
[tex]V_{R}=-3.759\times10^{-4}\ V[/tex]
The electric potential at 2.81 cm is [tex]-3.759\times10^{-4}\ V[/tex].
Hence, This is the required solution.
