Compute voltage inside sphere of uniform charge

AI Thread Summary
To find the electric potential inside and outside a uniformly charged solid sphere, the electric field outside the sphere is derived as E = (q)/(4πε₀r²), leading to a potential V = q/(4πε₀r). For the inside of the sphere, the electric field is not zero and is calculated using Gauss's law, where the charge enclosed is proportional to the volume within the radius r. The charge density is given by ρ = q/(4/3 π R³), and the electric field inside the sphere can be expressed as E = (q/(4/3 π R³))(r)/(ε₀). The challenge lies in correctly applying these principles to derive the electric field and potential inside the sphere.
Gary Roach
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Homework Statement


Problem 2.21 from Introduction to Electrodynamics, David J. Griffiths, Third Edition.

Find the potential inside and outside a uniformly charged solid sphere who's radius is R and whose total charge is q. Use infinity as your reference point.


Homework Equations


Given: q, R, r'(r inside sphere)
Variable: r
∫E da =q/ε0
V = -∫^{r}_{∞}E dl


The Attempt at a Solution


Outside of sphere:
|E|∫s r2sinΘdΘdø = (1/ε0)q
E = (q)\widehat{r}/(4π εor2)
V = -∫^{r}_{∞}E dr =q/4πε0r

The above is pretty straight forward. On the other hand, the Voltage inside the sphere completely illudes me. From the answer book, I know that is an A - B type problem but can't seem to get my mind around the concept. I know that the Voltage is a function of the radius but the E field is not zero as in a hollow sphere. How is the basic equation for the E field inside the sphere derived?
 
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charge inside the gaussian surface at r<R is proportional to the Volume enclosed: rho V_inside
where rho = q/(4/3 pi R^3)
 
For a radius r>R, consider the sphere with a smaller radius (smaller than r) and the hollow sphere with a larger radius (larger than r) as separate objects. How can you calculate their field contributions?
 
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