Calculation of electric field from a set of equipotential surfaces

Click For Summary
SUMMARY

The discussion focuses on calculating the electric field from a set of concentric hemispherical equipotential surfaces with varying potential differences. The electric field is determined to be radial and directed along the negative gradient of the potential, expressed mathematically as E = -dV/dr. The solution involves deriving the relationship between potential (V) and radial distance (r) and applying a radial line integral of the electric field to equate it to the potential difference. The participant successfully solved the problem using these insights.

PREREQUISITES
  • Understanding of electric fields and equipotential surfaces
  • Familiarity with the concept of electric potential and its gradient
  • Knowledge of Gaussian surfaces and their applications in electrostatics
  • Ability to perform line integrals in vector calculus
NEXT STEPS
  • Study the derivation of the relationship between electric potential and radial distance in spherical coordinates
  • Learn about the application of Gauss's Law in electrostatics
  • Explore the mathematical techniques for performing line integrals in vector fields
  • Investigate the behavior of electric fields in non-uniform charge distributions
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, electrical engineers, and anyone involved in solving problems related to electric fields and potentials.

kihr
Messages
98
Reaction score
0

Homework Statement


A set of concentric hemispherical surfaces is given, each of which is an equipotential surface. These concentric surfaces do not, however, have the same value of potential, and the potential difference between any two surfaces is also not constant. The surfaces are spaced apart at the same radial distance between each other. I need to find the direction and value of the electric field in this region.


Homework Equations


The electric field will be in the radial direction as the field has to be normal to the equipotential surface at each point. Its direction will be along the negative gradient of the potential (E = - dV/dr).


The Attempt at a Solution



I need some clues as to how to calculate the value of the electric field. This would have been easy if the relation between V and r were known. I guess this needs to be derived. Please give me some hints as to how to tackle this. Thanks.
 
Physics news on Phys.org
take a guassian surface with no charge inside to convince yourself the field decreases with 1/r^2, between any 2 plates

then take a radial line integral of the field, which must be equal to the potential difference to solve for E
 
With your tips I was able to solve the problem. Many thanks.
 

Similar threads

Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
5K
Replies
26
Views
6K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
9
Views
2K