Gradient, potential and electric field

AI Thread Summary
The discussion revolves around calculating the electric field and potential for a sphere with a specific charge density using Gauss’ law. The electric field inside the sphere is E = 1/ε0, while outside it is E = a^2/(ε0*r^2). For potential calculations, the potential outside the sphere is V(r) = -a^2/(ε0*r) and inside is V(r) = -2a/ε0. The potential at the origin is set to zero, which affects the calculations in part c). The final electric field derived from the gradient of the potential matches the initial findings from Gauss' law, confirming the calculations.
brad sue
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Hi ,
I need help about procedure on this problem:

Consider a sphere of charge density ρv= 2/r[C/m3] having a radius a (no charge outside), centered at the origin. Perform the following steps:
a) Calculate the electric field both inside and outside the sphere using Gauss’ law.
b) Calculate the potential inside and outside the sphere by integrating the electric field to get
the potential, assuming zero Volts at infinity.
c) Calculate the potential assuming that the reference is at the origin zero, V(0,0,0) =0
d) Find the electric field inside and outside the sphere by taking the gradient of the potential.
You should be able to verify that your answer is the same as the electric field that you started with from Gauss’s law.


I found the first question:
for r<a -> E=1/ε0

for r>a--> E=a^2/(ε0*r^2)

Now for the second question b)

for r>a :
V(r)-V(inf)=- ∫(a^2/(ε0*r^2)dr) [from infinity to r]=[a^2/(ε0*r)]+ (2a/(ε0))

for r<a:
V(r)-V(a)=- ∫(1/(ε0)dr) [from a to r]=-r/ε0

I did the same approach with the question (but not so sure about the method) c).

Now when I want to find that electric field with the gradient, I don't find the same answer of E I started with.

Perhaps my method is wrong.
Please,please can someone show me How to do this problem?
At least one part to show me how we verify the electric field with the gradient [especially with V(r=0)=0].

Thank you
 
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for your help!</code>For part b), your answer is correct.For part c), you need to subtract the potential at the origin from your answer in part b). Since V(r=0) = 0, this means that the potential inside the sphere is equal to -2a/ε0 and the potential outside is equal to -a^2/(ε0*r). For part d), you can calculate the electric field inside and outside the sphere by taking the gradient of the potential from part c). Inside, the electric field is E = 1/ε0 and outside, the electric field is E = a^2/(ε0*r^2). This is the same as the electric field you obtained in part a).
 
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