1) Electric potential inside the sphere: ![\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7B8%5Cpi%20%5Cepsilon_0%20R%7D%283-%5Cfrac%7Br%5E2%7D%7BR%5E2%7D%29)
2) Ratio Vcenter/Vsurface: 3/2
3) Find graph in attachment
Explanation:
1)
The electric field inside the sphere is given by
![E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%5Cfrac%7BQr%7D%7BR%5E3%7D)
where
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,
![E(r)=-\frac{dV(r)}{dr}](https://tex.z-dn.net/?f=E%28r%29%3D-%5Cfrac%7BdV%28r%29%7D%7Bdr%7D)
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,
![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)](https://tex.z-dn.net/?f=V%28R%29-V%28r%29%3D-%5Cint%5Climits%5ER_r%20%20E%28r%29dr%3D-%5Cfrac%7BQ%7D%7B4%5Cpi%20%5Cepsilon_0%20R%5E3%7D%5Cint%20r%20dr%20%3D%20%5Cfrac%7B-Q%7D%7B8%5Cpi%20%5Cepsilon_0%20R%5E3%7D%28R%5E2-r%5E2%29)
The potential at the surface, V(R), is that of a point charge, so
![V(R)=\frac{Q}{4\pi \epsilon_0 R}](https://tex.z-dn.net/?f=V%28R%29%3D%5Cfrac%7BQ%7D%7B4%5Cpi%20%5Cepsilon_0%20R%7D)
Therefore we can find the potential inside the sphere, V(r):
2)
At the center,
r = 0
Therefore the potential at the center of the sphere is:
![V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}](https://tex.z-dn.net/?f=V%28r%29%3D%5Cfrac%7BQ%7D%7B8%5Cpi%20%5Cepsilon_0%20R%7D%283-%5Cfrac%7Br%5E2%7D%7BR%5E2%7D%29%5C%5CV%280%29%3D%5Cfrac%7B3Q%7D%7B8%5Cpi%20%5Cepsilon_0%20R%7D)
On the other hand, the potential at the surface is
![V(R)=\frac{Q}{4\pi \epsilon_0 R}](https://tex.z-dn.net/?f=V%28R%29%3D%5Cfrac%7BQ%7D%7B4%5Cpi%20%5Cepsilon_0%20R%7D)
Therefore, the ratio V(center)/V(surface) is:
![\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BV%280%29%7D%7BV%28R%29%7D%3D%5Cfrac%7B%5Cfrac%7B3Q%7D%7B8%5Cpi%20%5Cepsilon_0%20R%7D%7D%7B%5Cfrac%7BQ%7D%7B4%5Cpi%20%5Cepsilon_0%20R%7D%7D%3D%5Cfrac%7B3%7D%7B2%7D)
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
, as found in part b) (where
)
- Between r and R, the potential decreases as ![-\frac{r^2}{R^2}](https://tex.z-dn.net/?f=-%5Cfrac%7Br%5E2%7D%7BR%5E2%7D)
- Then at r = R, the potential is ![V(R)](https://tex.z-dn.net/?f=V%28R%29)
- Between r = R and r = 3R, the potential decreases as
, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (
)
Learn more about electric fields and potential:
brainly.com/question/8960054
brainly.com/question/4273177
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