To solve this problem we will start by applying the given load ratio, and we will rely on the two types of distances given. Later we will use Gauss's law and through its integrals, in which it is equivalent to the potential we will obtain its value in the center of the sphere. Since it is uniformly charged we have to,
[tex]\frac{Q'}{Q} = \frac{4\pi r^3}{4\pi R^3}[/tex]
[tex]Q' = \frac{r^3}{R^3} Q[/tex]
By Gauss Law
[tex]\phi = E \cdot 4\pi r^2[/tex]
Here, E is the electric Field and is equal to
[tex]E = \frac{Q'}{\epsilon_0}[/tex]
For [tex]\epsilon_0[/tex] being the Permeability constant at free space
Replacing with the previous value we have,
[tex]\phi = \frac{Qr^3}{\epsilon_0 R^3}[/tex]
Then the value of the electric field is,
[tex]E = \frac{QR}{4\pi \epsilon_0 R^3}[/tex]
Now potential
[tex]V = \int_0^R E\cdot dr'[/tex]
[tex]V = - \frac{QR^2}{8\pi \epsilon_0 R^3}[/tex]
[tex]V = -\frac{Q}{8 \pi \epsilon_0 R}[/tex]
The electric potential on the surface of the sphere is found by using Gauss' theorem. The value of electric potential is obtained to be
[tex]V=-\frac{Q}{8 \pi \epsilon_0 R}[/tex].
Given the uniformly charged sphere has a charge 'Q' and radius 'R'.
The volume charge density of the sphere is given by;
[tex]\rho = \frac{Q}{V} =\frac{Q}{4\pi R^3}[/tex]
At a distance 'r' from the centre of the sphere, the charge present can be given as;
[tex]q=\rho V=\frac{Q}{4 \pi R^3} \times 4\pi r^3=\frac{Qr^3}{R^3}[/tex]
According to Gauss theorem;
[tex]\epsilon _0\, \phi=q_{\,enclosed}[/tex]
Where, [tex]\phi[/tex] is the electric flux.
Also, [tex]\phi=EA=E\times 4\pi r^2[/tex]
Substituting for [tex]\phi[/tex] and 'q' in Gauss theorem, we can find the electric field at the point 'r';
[tex]\epsilon_0 \times E \times 4 \pi r^2 =\frac{Qr^3}{R^3}[/tex]
[tex]\implies E =\frac{Qr^3}{4 \pi \epsilon_0 r^2R^3}=\frac{Qr}{4 \pi \epsilon_0 R^3}[/tex]
Now we know that the potential at the point R is given by;
[tex]V=-\int\limits^R_0 {E} \, dr = -\int\limits^R_0 {\frac{Qr}{4 \pi \epsilon_0 R^3} } \, dr=-[ {\frac{Q}{4 \pi \epsilon_0 R^3} }\times \frac{r^2}{2} ]^R_0=-\frac{Q}{8 \pi \epsilon_0 R}[/tex]
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