Gen Phys 2

Electric potential

Electric potential energy

\begin{equation} W_{a \to b} = U_a - U_b \end{equation} \begin{equation} U = \frac{1}{4 \pi \epsilon_0} \frac{q_{\textnormal{  test}} \, \, q_{\textnormal{  source}}}{r} \end{equation} \begin{equation} U = \frac{q_{\textnormal{  source}}}{4 \pi \epsilon_0} \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} + \frac{q_3}{r_3} \, + \, ... \right) \end{equation} \begin{equation} = \frac{q_{\textnormal{  source}}}{4 \pi \epsilon_0} \sum_{i = 1} \frac{q_i}{r} \end{equation}

Electric potential

\begin{equation} V = \frac{U}{q_0} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} \end{equation} \begin{equation} V = \frac{U}{q_0} = \frac{1}{4 \pi \epsilon_0} \sum_{i = 1} \frac{q_i}{r} \end{equation} \begin{equation} V = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{r} \end{equation} \begin{equation} V_a - V_b = \int_a^b \vec{E} \cdot d\vec{l} = \int_a^b E \textnormal{ cos } \gamma \, dl \end{equation}

Calculating electric potential

Potential due to a charged particle

Potential due to an electric dipole

Potential due to a continuous charge distribution

Equipotential surfaces

Potential of a charged isolated conductor

Electric field from electric potential

\begin{equation} E_x = - \frac{\partial V}{\partial x} \quad E_y = - \frac{\partial V}{\partial y} \quad E_z = - \frac{\partial V}{\partial z} \end{equation} \begin{equation} \vec{E} = - \left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \end{equation}