1) Electric potential inside the sphere: [tex]\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})[/tex]
2) Ratio Vcenter/Vsurface: 3/2
3) Find graph in attachment
Explanation:
1)
The electric field inside the sphere is given by
[tex]E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}[/tex]
where
[tex]\epsilon_0=8.85\cdot 10^{-12}F/m[/tex] is the vacuum permittivity
Q is the charge on the sphere
R is the radius of the sphere
r is the distance from the centre at which we compute the field
For a radial field,
[tex]E(r)=-\frac{dV(r)}{dr}[/tex]
Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,
[tex]V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)[/tex]
The potential at the surface, V(R), is that of a point charge, so
[tex]V(R)=\frac{Q}{4\pi \epsilon_0 R}[/tex]
Therefore we can find the potential inside the sphere, V(r):
[tex]V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})[/tex]
2)
At the center,
r = 0
Therefore the potential at the center of the sphere is:
[tex]V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}[/tex]
On the other hand, the potential at the surface is
[tex]V(R)=\frac{Q}{4\pi \epsilon_0 R}[/tex]
Therefore, the ratio V(center)/V(surface) is:
[tex]\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}[/tex]
3)
The graph of V versus r can be found in attachment.
We observe the following:
- At r = 0, the value of the potential is [tex]\frac{3}{2}V(R)[/tex], as found in part b) (where [tex]V(R)=\frac{Q}{4\pi \epsilon_0 R}[/tex])
- Between r and R, the potential decreases as [tex]-\frac{r^2}{R^2}[/tex]
- Then at r = R, the potential is [tex]V(R)[/tex]
- Between r = R and r = 3R, the potential decreases as [tex]\frac{1}{R}[/tex], therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ([tex]\frac{1}{3}V(R)[/tex])
Learn more about electric fields and potential:
brainly.com/question/8960054
brainly.com/question/4273177
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