1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 at 18:11
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.