Electric Potential

Electric Potential¶

\[\Delta V = V_b - V_a = -\frac{W_{ab}}{q_0} = -\int_a^b \vec{E} \cdot d\vec{l}\]

\[V_p = -\int_{\infty}^p E \cdot dl = \int_p^{\infty} E \cdot dl\]

The electric potential around a point charge is \(\(V(r) = \frac{q}{4\pi\epsilon_0 r}\)\)

The electric potential around a electric dipole is \(\(V(r) = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r_1} - \frac{q}{r_2} \right) = \frac{q}{4\pi\epsilon_0} \frac{r_2 - r_1}{r_1r_2} \approx \frac{p\cos\theta}{4\pi\epsilon_0 r^2}\)\)

For electric quadrupole, the electric potential on the axis of the quadrupole is \(\(V(r) = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r - d} + \frac{-2q}{r} + \frac{q}{r + d} \right) = \frac{1}{4\pi\epsilon_0} \frac{2qd^2}{r(r^2 - d^2)}\)\)

When \(r >> d\), we have \(V(r) = \dfrac{Q}{4\pi\epsilon_0 r^3}\).

Relation between Electric Field and Potential¶

\[ \begin{gathered} E\cos\theta = -\frac{dV}{dl} \\\\ \Rightarrow E = -\nabla V \end{gathered} \]

In cartesian coordinates, \(\(\nabla \vec{V} = \frac{\partial V}{\partial x}\hat{x} + \frac{\partial V}{\partial y}\hat{y} + \frac{\partial V}{\partial z}\hat{z}\)\)

In spherical coordinates, \(\(\nabla \vec{V} = \frac{\partial V}{\partial r}\hat{r} + \frac{1}{r} \frac{\partial V}{\partial \theta}\hat{\theta} + \frac{1}{r\sin\theta} \frac{\partial V}{\partial \phi}\hat{\phi}\)\)

In cylindrical coordinates, \(\(\nabla \vec{V} = \frac{\partial V}{\partial r}\hat{r} + \frac{1}{r} \frac{\partial V}{\partial \theta}\hat{\theta} + \frac{\partial V}{\partial z}\hat{z}\)\)

When \(r >> a\), the electric field around a dipole is

\[ \begin{gathered} V(r, \theta) = \frac{1}{4\pi\epsilon_0} \frac{p\cos\theta}{r^2} \\\\ \Rightarrow E = - \frac{\partial V}{\partial r} - \frac{1}{r} \frac{\partial V}{\partial \theta} = \frac{p}{4\pi\epsilon_0 r^3}((2\cos\theta)\hat{r} + (\sin\theta)\hat{\theta}) \end{gathered} \]

The electric field along the axis of a charged disk is

\[ \begin{gathered} dV = \frac{dq}{4\pi\epsilon_0\sqrt{z^2 + r^2}} = \frac{2\pi rdr \cdot \sigma}{4\pi\epsilon_0\sqrt{z^2 + r^2}} \\\\ \Rightarrow V = \frac{\sigma}{2\epsilon_0} \int_0^R \frac{rdr}{\sqrt{z^2 + r^2}} = \frac{\sigma}{2\epsilon_0} (\sqrt{z^2 + R^2} - z) \\\\ \Rightarrow E = - \frac{\partial V}{\partial z} = \frac{\sigma}{2\epsilon_0} (1 - \frac{1}{\sqrt{1 + (R/z)^2}}) \end{gathered} \]