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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.

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

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

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

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  • $\begingroup$ 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

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