Gradient, potential and electric field

[brad sue]

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

 

[Nate810]

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).

 

[m2003]

for your question. I will provide a step-by-step response to help guide you through the problem.

a) To calculate the electric field using Gauss' law, we can use the formula E = Q/ε0, where Q is the total charge enclosed by a Gaussian surface and ε0 is the permittivity of free space. In this case, our Gaussian surface will be a sphere of radius r.

For r < a (inside the sphere), the total charge enclosed is simply the charge density multiplied by the volume of the sphere, which gives us Q = (4/3)πr^3ρv. Substituting this into the formula, we get E = (4/3)πr^3ρv/ε0.

For r > a (outside the sphere), the total charge enclosed is the total charge of the sphere, which is (4/3)πa^3ρv. Substituting this into the formula, we get E = (4/3)πa^3ρv/ε0r^2.

b) To calculate the potential, we can use the formula V = -∫E dr, where E is the electric field and dr is an infinitesimal displacement.

For r > a, we can integrate the electric field expression we found in part a) to get V = (2/3)πa^3ρv/ε0r + C, where C is a constant of integration. We can set C = 0, since the potential at infinity is assumed to be zero.

For r < a, we can integrate the electric field expression we found in part a) to get V = (4/3)πr^3ρv/ε0 + C. We can set C = -V(a), where V(a) is the potential at the surface of the sphere. This is because we know that the potential at the surface of the sphere is equal to the potential inside the sphere.

c) To set the reference potential at the origin, we can simply subtract the potential at the origin from the potential we found in part b). Since we set the potential at the origin to be zero, we get V(r) = (2/3)πa^3ρv/ε0r - (2/3)πa^3ρv/ε0r = 0.

d) To find the electric field from