# Relation between electric potential and Electric field

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.

• @AaronStevens why V=(E).d for constant (E). – 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

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

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

• 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. – Cryo Sep 2 at 2:20
• Thank you for helping me. – user359206 Sep 2 at 4:06
• @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. – Aaron Stevens Sep 2 at 4:27
• 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 – user359206 Sep 2 at 4:58
• @user359206 You should be able to accept answers on your own questions, which should let you get some rep points – Aaron Stevens Sep 2 at 5:01