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This is an exercise to calculate the difference in potential between two points $A$ and $B$ in an uniform $\vec{E}$ electric field.

It goes like $V_{B} - V_{A} = -\int_{A}^{B} \vec{E} \cdot d \vec{s} = -E \cdot d$. I get this, but I don't understand why assume that the potential in $B$ is lower than in $A$.

Can someone explain? Thanks.

Difference of Potential in an uniform field

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  • $\begingroup$ Wouldn't that be VA - VB? I mean, according to the diagram, and you, potential in B is lower than in A. So, higher potential - lower potential. Also, the limits in the integral. $\endgroup$ – Tapi Apr 28 at 19:48
  • $\begingroup$ The definition given to me was $ V_{B} - V_{A} = - \int_{A}^{B} \vec{E} \cdot d \vec{s}$. Potential B lower than A is not a given, it's expected for me to work it out. $\endgroup$ – Leon Held Apr 28 at 19:54
  • $\begingroup$ @Tapi $d\vec{s}$ is directed from A to B. The direction of the path is indicated by the bounds of the path integral. $\endgroup$ – G. Smith Apr 28 at 21:15
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I get this, but I don't understand why assume that the potential in 𝐵 is lower than in 𝐴.

The direction of the electric field lines is the direction of the force a positive charge would experience if placed in the field. In order to move a positive charge from B to A which is against the field external work must be done on the charge. This increases its potential energy. If the positive charge is placed at A it will naturally move from A to B, and therefore loses electrical potential energy. The analogy with a gravitational field and gravitational potential energy may be helpful.

In the figure below I have rotated your diagram. Imagine now that instead of the lines being an electric field, that they represent a uniform gravitational field. Like our positive electric charge, the gravitational field lines point in the direct of the force that an object would experience if placed in the gravitational field.

If you raise an object from B to A, against the gravitational field, you know you have to do work against the gravitational force and that the object gains gravitational potential energy. If you release an object from rest at A it falls losing potential energy and gaining kinetic energy. The point B is said to be at a lower gravitational potential than A. It's the same case for electrical potential. The point B is at a lower electrical potential then the point A.

Hope this helps.

enter image description here

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  • $\begingroup$ Your explanation plus Charles Hudgins integral solving nailed it for me. Thank you very much! $\endgroup$ – Leon Held Apr 28 at 22:18
  • $\begingroup$ @Leon Held you’re welcome. Many times more than one answer provides more help than one. I liked Charles Hudgins answer too and upvoted it. $\endgroup$ – Bob D Apr 28 at 22:26
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"But I don't understand why to assume that the potential in B is lower than in A."

The direction of the electric field intensity is always from higher potential to lower potential. So, according to the diagram, point B being to the right (compared to point A) has a lower potential than A.

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For a static electric field, we traditionally define the potential as $$V(\mathbf{x}) = - \int_\mathbf{a}^\mathbf{x} \mathbf{E} \cdot d\mathbf{s}$$ where $\mathbf{a}$ is some fixed point. This means $$V(B) - V(A) = - \int_\mathbf{a}^B \mathbf{E} \cdot d\mathbf{s} + \int_\mathbf{a}^A \mathbf{E} \cdot d\mathbf{s} = - \int_\mathbf{a}^B \mathbf{E} \cdot d\mathbf{s} - \int_A^\mathbf{a} \mathbf{E} \cdot d\mathbf{s} = -\int_A^B \mathbf{E} \cdot d\mathbf{s}$$ If $$\mathbf{E} = \frac{E(B - A)}{||B - A||}$$ and we choose $$\gamma(t) = tB + (1-t)A$$ as our path of integration, we find \begin{align*}V(B) - V(a) &= -\int_A^B \mathbf{E} \cdot d\mathbf{s} \\&= -\int_0^1 \frac{E(B - A)}{||B - A||} \cdot \gamma'(t) dt \\&= -\int_0^1 \frac{E(B - A)}{||B - A||} \cdot (B - A) dt \\&= -\int_0^1 E||B - A|| dt = - E||B - A|| \end{align*} As you can see, there were no assumptions made. We applied the definition of the potential and computed an integral.

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