9.2 Electric Potential

Just like it was useful to develop the concept of electric fields to represent the electric force irrespective of the test charge, the electric potential ${V}$ extends the electric potential energy ${U}$ to be irrespective of the test charge. The potential difference ${\Delta V}$ (also known as voltage) between points a and b is defined by ${\Delta V = \int_{a}^{b} -\vec{E} \cdot d\vec{s}}$ where ${\vec{E}}$ is the electric field acting on the test charge. We use the same process that we used for finding the potential energy to find the electric potential of a charge at point a. ${\int_{\infty}^{a} -\vec{E} \cdot d\vec{r} = \int_{a}^{\infty} \vec{E} \cdot d\vec{r} = \int_{a}^{\infty} \frac{q_1}{4\pi\epsilon_0 r^2} \hat{r} \cdot d\vec{r} = \frac{q_1}{4\pi\epsilon_0 a} = \frac{kq_1}{a}}$.

We can use the superposition principle to find the potential of a system of charges: ${V = \Sigma \frac{q_i}{4\pi\epsilon_0 r_i} = \Sigma \frac{kq_i}{r_i}}$. When the charge is continuous we integrate over all the charges to get ${V = \int_{0}^{Q} \frac{dq}{4\pi\epsilon_0 r}}$. The electric potential is a scalar, unlike the electric field. This means that when we evaluate the potential of continuous distributions, we don't have to worry about the vectors pointing in different directions. This means that for the majority of cases, it is significantly easier to evaluate the electric potential due to a distribution compared to the electric field.

Because of this fact, one technique for evaluating the electric field of a continuous distribution is evaluating the electric potential and then taking the derivative along each axis to get the electric field, ${\vec{E} = -\frac{dV}{dx} \hat{i} - \frac{dV}{dy} \hat{j} - \frac{dV}{dz} \hat{k}}$. Equipotential surfaces are a way to visualize the electric potential.