Griffiths Electrodynamics book: Electric potential

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Homework Help Overview

This discussion revolves around electric potential in the context of point charges and the implications of changing one charge from positive to negative. The original poster references a specific problem from Griffiths' Electrodynamics, focusing on the potential at a point above the midpoint of two charges and the resulting electric field considerations.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between electric potential and electric field, questioning how potential information translates to field components. There are discussions on the implications of knowing potential only along the z-axis and how that affects the calculation of the electric field in other directions.

Discussion Status

The discussion is ongoing, with participants providing insights into the limitations of using the potential at a specific point to derive the electric field. Some participants suggest that a broader understanding of the potential function is necessary to compute the electric field accurately, while others are seeking clarification on the relationship between scalar potential and vector fields.

Contextual Notes

There is a mention of symmetry in electric fields and the need for a general potential function to determine the electric field direction. Additionally, a separate problem involving infinite parallel plates and varying charge density is introduced, but it is advised to start a new thread for that topic.

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Homework Statement



This is from Prob. 2.25
Two point charges with separation d, P is a point at a distance z above the mid-point of the charges.
The last sentence asked if one of the positive charges is changed to a negative one, what is the potential at P? What field does it suggest? Explain the discrepancy.

Homework Equations


E = -grad V

The Attempt at a Solution


V=0 obviously
E = -grad V = 0
What is the reason behind that gives this result?
 
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Remember, -\vec{\nabla}V=\frac{\partial V}{\partial x}\hat{x}+\frac{\partial V}{\partial y}\hat{y}+\frac{\partial V}{\partial z}\hat{z} Since you only know V on the z-axis, you cannot possibly calculate \frac{\partial V}{\partial x} and \frac{\partial V}{\partial y}}. Clearly, any E-field will point in the x-direction, and so it is necessary to determine V(x,y,z) at points off of the z-axis to find \frac{\partial V}{\partial x} and hence E.
 
Last edited:
I had exactly the same question and I kind of understand what you are saying but since V is scalar and E is vector, isn't V supposed to possesses all the information that the three components of E possess? Would you please suggest me how you would compute E = -\nabla V in this case and compare it with calculation using Gauss's Law?

Thanks
 
The equation \vec{E}=-\vec{\nabla}V applies to V(x,y,z) (in Cartesian coordinates anyways). The potential you've calculated is actually V(0,0,z) (the -potential on the z-axis) and so you do not know how V varies with x or y, and you cannot use V(0,0,z) to compute E. If you wanted to compute E from the potential, then you would need to find the potential at a general point (x,y,z) (or even just a point on the x-axis in this case) first and use that potential.

In some cases, you know from symmetry that E points in the z direction and so knowing the functional dependence of V(z) is enough. In this case however, E points in the x direction and so you need to know the functional dependence of V(x).
 
Thanks,

I also have another question...
Two infinite parallel plates separated by a distance s are at potential 0 and V_0
a) Use Poisson's eqn to find potential V in the region between the plates where the space charge density is rho = rho_0(x/s). The distance x is measured from the plate at 0 potential.
b)What are the charge densities in the plate?

For this problem, I started with number of ways but none of them seem to be working.
 
You should start a new thread for that problem.
 

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