I'm unable to understand why the integration of $\mathbf E$ is done in two different ways for constant $\mathbf E$ and varying $\mathbf E$, as in case of parallel plate capacitor and spherical capacitor. More specifically, why do we get $\Delta V=Ed$ for the constant field?

Also, if somehow I understand the idea of integration described above, it doesn't seem to work for calculating the potential energy $U$ for the varying force.

  • $\begingroup$ @AaronStevens why V=(E).d for constant (E). $\endgroup$ – user359206 Sep 1 at 3:41

In general, the difference in electric potential between position $\mathbf a$ and $\mathbf b$ is given by $$\Delta V=V(\mathbf b)-V(\mathbf a)=-\int_{\mathbf a}^{\mathbf b}\mathbf E\cdot\text d\mathbf r$$ where the integral is a line integral following any path from $\mathbf a$ to $\mathbf b$. This definition is true for any static electric field, constant or not.

However, if the field is constant along the integration path, then we are allowed to take the $\mathbf E$ term outside of the integral: $$\Delta V=V(\mathbf b)-V(\mathbf a)=-\mathbf E\cdot\int_{\mathbf a}^{\mathbf b}\text d\mathbf r$$

Now, the line integral of $\text d\mathbf r$ is just the vector that points from the start to the end of the path, i.e. $$\int_{\mathbf a}^{\mathbf b}\text d\mathbf r=\mathbf b-\mathbf a$$ Therefore, for a constant electric field, $$\Delta V=-\mathbf E\cdot(\mathbf b-\mathbf a)$$

Now, if you are only interested in the magnitude of the potential difference, and if the field points in the same direction as the displacement, we can simplify to $$|\Delta V|=|\mathbf E|d$$ Where $d$ is the magnitude of $\mathbf b-\mathbf a$.

This same idea is true for electric potential energy and forces as well, because potential and field are just the energy and force respectively per unit charge, i.e. $\Delta V=\Delta U/q$ and $\mathbf E=\mathbf F/q$. So if you take the above section and divide everything by $q$ then you are good to get

$$\Delta V/q=V(\mathbf b)/q-V(\mathbf a/q)=-\int_{\mathbf a}^{\mathbf b}\mathbf E/q\cdot\text d\mathbf r$$

$$\Delta U=U(\mathbf b)-U(\mathbf a)=-\int_{\mathbf a}^{\mathbf b}\mathbf F\cdot\text d\mathbf r$$

  • $\begingroup$ Nice answer. I would only add that in electrostatics, $\boldsymbol{\nabla}.\mathbf{E}=\rho/\epsilon$ and $\boldsymbol{\nabla}\times\mathbf{E}=\mathbf{0}$, together with Helmholtz decomposition (en.wikipedia.org/wiki/Helmholtz_decomposition) imply that $\mathbf{E}=-\boldsymbol{\nabla} V$ for some scalar function $V$, given suitable boundary conditions. The rest is integration as you describe. $\endgroup$ – Cryo Sep 2 at 2:20
  • $\begingroup$ Thank you for helping me. $\endgroup$ – user359206 Sep 2 at 4:06
  • $\begingroup$ @user359206 Of course. Glad I could help. Remember to upvote all useful answers. Also, make sure to select an answer as the accepted answer for future readers. $\endgroup$ – Aaron Stevens Sep 2 at 4:27
  • $\begingroup$ I'm new here and don't know much how it works when I upvote it shows I don't have enough badges to do so $\endgroup$ – user359206 Sep 2 at 4:58
  • $\begingroup$ @user359206 You should be able to accept answers on your own questions, which should let you get some rep points $\endgroup$ – Aaron Stevens Sep 2 at 5:01

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.