Respuesta :
Explanation:
The given data is as follows.
Radius of the sphere (R) = 10 cm = 0.01 m (as 1 m = 100 cm)
Distance from the center (r) = 4 cm = 0.04 m
Charge density ([tex]\sigma[/tex]) = 100 [tex]nC/m^{3}[/tex]
= [tex]100 \times 10^{-9} C/m^{3}[/tex] (As 1 nm = [tex]10^{-9} m[/tex])
As the relation between charge and potential difference is as follows.
Q = [tex]\sigma V[/tex]
= [tex]\sigma (\frac{4}{3} \pi r^{3})[/tex]
= [tex]100 \times 10^{-9} C/m^{3} \times \frac{4}{3} \pi (0.01)^{3}[/tex]
= [tex]4.19 \times 10^{-10} C[/tex]
Expression for electric field is as follows.
E(r) = [tex]k \frac{qr}{R^{3}}[/tex]
Electric potential, V(r) = [tex]-\int_{0}^{r} E(r) dr[/tex]
= [tex]-\int_{0}^{r} k \frac{qr}{R^{3}} dr[/tex]
= [tex]-k (\frac{qr^{2}}{2R^{3}})[/tex]
= [tex]-9 \times 10^{9} Nm^{2}/C^{2} (\frac{4.19 \times 10^{-10} C (0.04)^{2}}{2(0.1)^{3}})[/tex]
= -3 V
Thus, we can conclude that the magnitude of the potential difference between the center and a point 4.0 cm away is -3 V.
