Gradient, potential and electric field

Click For Summary
SUMMARY

The discussion focuses on calculating the electric field and potential of a sphere with a charge density of ρv = 2/r C/m³ and radius a, using Gauss' law. The electric field inside the sphere is E = 1/ε0, while outside it is E = a²/(ε0*r²). The potential is derived by integrating the electric field, resulting in V(r) for r > a as V(r) = -a²/(ε0*r) + (2a/ε0) and for r < a as V(r) = -r/ε0. The gradient of the potential confirms the electric field values obtained from Gauss' law.

PREREQUISITES
  • Understanding of Gauss' law in electrostatics
  • Knowledge of electric field and potential concepts
  • Familiarity with integration techniques in physics
  • Basic understanding of gradients in vector calculus
NEXT STEPS
  • Study the application of Gauss' law for different charge distributions
  • Learn about electric potential and its relationship with electric fields
  • Explore vector calculus, specifically gradient operations
  • Investigate the implications of reference points in potential calculations
USEFUL FOR

Students and professionals in physics, electrical engineering, and anyone studying electrostatics, particularly those working with electric fields and potentials in spherical charge distributions.

brad sue
Messages
270
Reaction score
0
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
 
Physics news on Phys.org
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).
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
470
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
Replies
10
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K