The relation is : $$\text dV = -\mathbf E\cdot \text d\mathbf R$$ (both of $E$ and $\text d\mathbf R$ are vectors ) . I have seen some questions and derivations where it is assumed constant. Please correct me if I am wrong.
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$\begingroup$ In various classical cases $E$ is not constant. For example, when we calculate the potential in the space around a point-like particle or a spherical charged particle. I would say, $E$ is constant only when there are flat plates or distributions of charges. Maybe you refer to the direction of $E$. Please clarify, or put a link to an example. $\endgroup$ – Doriano Brogioli May 1 '19 at 18:11
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This is true in general. You can start with the definition of potential energy of the electrostatic force (or really any conservative force in general). $$\mathbf F=-\nabla U$$ And then divide both sides by the charge exerting the field in question: $$\frac{\mathbf F}q=-\nabla \left(\frac Uq\right)$$
The value on the right is just the electric field, while the value on the left is the negative gradient of the electric potential $$\mathbf E=-\nabla V$$
Or, using the fundamental theorem of calculus: $$V(\mathbf b)-V(\mathbf a)=\Delta V=\int_C\nabla V\cdot \text d\mathbf r=-\int_C\mathbf E\cdot\text d\mathbf r$$ where we have done a line integral along some path $C$ from some position $\mathbf r = \mathbf a$ to another position $\mathbf r = \mathbf b$
If we make $\mathbf a$ and $\mathbf b$ infinitesimally close so that $\mathbf b=\mathbf a+\text d\mathbf r$ then $$V(\mathbf a+\text d\mathbf r)-V(\mathbf a)=\text dV=-\mathbf E\cdot\text d\mathbf r$$
Notice how we didn't assume anything about a constant electric field. This is true for any electric field
However things work out nicely if the field is constant in space, since $$\Delta V=-\int_C\mathbf E\cdot\text d\mathbf r=-\mathbf E\cdot\int_C\text d\mathbf r=-\mathbf E\cdot\Delta\mathbf r$$ and you no longer have to perform a line integral.
So comparing what we have, in general: $$\text dV=-\mathbf E\cdot\text d\mathbf r$$ and for constant electric field: $$\Delta V=-\mathbf E\cdot\Delta\mathbf r$$ which shows the power of calculus very nicely: on the infinitesimal scale we can essentially treat our functions as constant.
answered May 1 '19 at 18:09
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In order to derive the equation you quote, E might just as well be constant. That is because, over the small displacement, dR, that you are considering, a reasonably well-behaved E will not change significantly.
The equation then follows from the definition of work, since E is force per unit charge (one a test charge) and dV is the work done per unit charge (on a test charge). The negative sign arises because if force and displacement are in the same direction, the charge will lose electrical potential energy.
answered May 1 '19 at 18:12
