Respuesta :
Answer:
Vb = k Q / 2R
Explanation:
To calculate the electric potential we use the relationship with the electric potential
dV = E. ds
Therefore we must find the electric potential for the interior of the sphere, for which we can use Gauss's law.
Ф = E .dA = [tex]q_{int}[/tex] / ε₀
We define a Gaussian surface as a sphere of radius r, in this case the electric field lines and the radial lines of the sphere are parallel so the scalar product is reduced to the algebraic product
The area to a sphere is
A = 4π R²
E A = q_{int} /ε₀
Let's use constant charge density
ρ = q_{int} / V
V = 4/3 π r³
q_{int} = ρ 4/3 π r³
We substitute
E = (ρ 4/3 π r³) / 4π r² ε₀
E = ρ / 3ε₀ r
Now we can find electrical potential
.dV = - ρ / 3ε₀ ∫ r dr
Vb –Va = - ρ / 3ε₀ r² / 2
We evaluate between the lower limits r = R and the upper limit r = r
Vb –Va = ρ / 3ε₀ (R² - r²) / 2
We substitute the density value
ρ = Q / (4/3 π R³)
Vb- Va = Q (1 / 4πε₀) (R²-r²) / 2R³
Where
k = 1 / 4π ε₀
Vb-Va = k Q / 2R³ (R² - r²)
In the center of the sphere r = 0
Vb = k Q / 2R
Answer:
V(0)=3Q/(8*pi*ε0R)
Explanation:
To find the values of the potential inside the solid sphere we have to use
[tex]V(r)=-\int E \cdot dr[/tex]
[tex]E=\frac{Q}{4\pi \epsilon_0 r^2}\hat{r}[/tex]
and we have also to take the charge density of the sphere as
[tex]\rho=\frac{Q}{V}=\frac{Q}{4\pi R^3/3}=\frac{3Q}{4\pi R^3}[/tex]
we can integrate from r to R (the radius of the solid sphere)
[tex]V=-\int \frac{Q}{4\pi \epsilon_0 r^2}dr=-\int_r^R \frac{1}{3\epsilon_0}\rho rdr\\\\V=-\frac{\rho}{3\epsilon_0}(\frac{R^2}{2}-\frac{r^2}{2})=\frac{Q(3R^2-r^2)}{8\pi \epsilon_0R^3}\\\\V(0)=\frac{3QR^2}{8\pi \epsilon_0 R^3}=\frac{3Q}{8\pi \epsilon_0 R}[/tex]
hope this helps!!
