Does equal electric field imply equal potential?

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Homework Help Overview

The discussion revolves around the relationship between electric fields and electric potential, specifically in the context of a given electric field described by E = 18/R² in the radial direction. The original poster is tasked with finding the electric potential between two points, A and B, located on the z-axis at different distances from the origin.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the electric field being the same at two points and whether this indicates equal potential. Questions arise about the symmetry of the problem and how it affects potential calculations. The original poster attempts to compute the potential difference using integration and seeks clarification on the expression for the differential path element.

Discussion Status

Participants are actively engaging with the concepts, questioning the assumptions about equipotential points and discussing the implications of the electric field's dependence on radial distance. Some guidance is offered regarding the integration process and the interpretation of potential differences, but no consensus has been reached on the initial assumptions regarding equipotentiality.

Contextual Notes

There are discussions about the symmetry of the electric field and potential, as well as the definitions of potential difference in relation to the points A and B. The original poster also notes the need for clarity on the direction of integration and the interpretation of the results.

fishingspree2
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Homework Statement


Given the E field
E = 18/R2 R, R is the radial direction.
Find the electric potential between A and B where A is at +2m and B at -4m, both on the z axis.


The Attempt at a Solution


My question is, since E field depends only on R, the distance between the point and the origin, then E field is the same for point B at -4m and point C at +4m. Therefore they are equipotential. Therefore I can compute the potential difference between A at +2m and C at +4m. Is this reasoning correct?
 
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No, equal electric field does not imply equal electric potential. The electric field is the gradient of the electric potential: E = -∇V, or in one dimension, E = -dV/dr.

So, if the electric field strength and direction are the same at two points, it doesn't mean that V is the same at those two points, it just means that the derivative of V is the same at those two points.

Example: electric field between two large parallel plates of uniform surface charge, is the same everywhere between the two plates, but electric potential increases linearly as you move from one plate to the other.

(The approximately uniform gravitational field close to the surface of the Earth is a totally analogous situation. The gravitational field strength is the same at two points at two different heights, but the gravitational potential is definitely different between those two points.)
 
Hmm...what I said above was the most general answer. HOWEVER in this case, it seems like the problem posesses some symmetry. If dV/dr depends only on r, then V will depend only on r as well. So V is the same at all points that are equidistant from the origin.
 
cepheid said:
Hmm...what I said above was the most general answer. HOWEVER in this case, it seems like the problem posesses some symmetry. If dV/dr depends only on r, then V will depend only on r as well. So V is the same at all points that are equidistant from the origin.
Some questions...
I'm trying to compute V between two general points for the given E field, just to see what it will give and also for the sake of it.
We know that

Vb-Va = -\int_{A}^{B}\overrightarrow{E}\cdot d\overrightarrow{l}

any idea how I could express d\overrightarrow{l}...?
also, should d\overrightarrow{l} be from A to B or from B to A?
is Vb-Va called the potential difference between A and B or the potential difference between B and A?
 
fishingspree2 said:
Some questions...
I'm trying to compute V between two general points for the given E field, just to see what it will give and also for the sake of it.
We know that

Vb-Va = -\int_{A}^{B}\overrightarrow{E}\cdot d\overrightarrow{l}

any idea how I could express d\overrightarrow{l}...?
also, should d\overrightarrow{l} be from A to B or from B to A?
is Vb-Va called the potential difference between A and B or the potential difference between B and A?
I found that d\overrightarrow{l}=dR\overrightarrow{R}+Rd\phi \overrightarrow{\phi}+dz\overrightarrow{z}

so
V_{B}-V_{A}-\int_{A}^{B}\overrightarrow{E}\cdot d\overrightarrow{l}=-\int_{R_{A}}^{R_{B}}\frac{18}{R^{2}}dR=\frac{18}{R_{B}}-\frac{18}{R_{A}}
is that correct? in that case it seems my initial assumption that -4 and 4 are equipotential is correct since it depends only on R which is 4 in both cases.
is that the potential change when you go from B to A or is it the potential change when you go from A to B?
 
fishingspree2 said:
I found that d\overrightarrow{l}=dR\overrightarrow{R}+Rd\phi \overrightarrow{\phi}+dz\overrightarrow{z}

so
V_{B}-V_{A}-\int_{A}^{B}\overrightarrow{E}\cdot d\overrightarrow{l}=-\int_{R_{A}}^{R_{B}}\frac{18}{R^{2}}dR=\frac{18}{R_{B}}-\frac{18}{R_{A}}
is that correct? in that case it seems my initial assumption that -4 and 4 are equipotential is correct since it depends only on R which is 4 in both cases.
is that the potential change when you go from B to A or is it the potential change when you go from A to B?

Obviously V_B - V_A is the potential change when you go from A to B. You're taking the difference between the final value and the initial value to compute the change.
 

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