9.1 - Electric Potential Part I

9.1.1 - Introduction to Electric Potential and Potential Difference

From earlier units, we learned that the force of the attraction between two charged objects is represented by Coulomb's Law, and the electric field relationship $\mathbf{F}_{e} = q\mathbf{E}$ such that q is a test charge (charge of negligible mass which is under the influence of the electric field). In this relationship, the force of attraction represented by Coulomb's Law between charged objects is a conservative force such as gravity. In mechanics we learned that the work done by an external agent in an electric field is equal to $\mathbf{mgh}$, and te work done by the gravitational force is $\mathbf{-mgh}$.

Similarly, there is a potential energy when working with objects under the influence of an electric field. In mechanics, the derivation for net work was equal to $\mathbf{\int_a^b F dx}$, which is also equivalent to the change in kinetic energy $\mathbf{\Delta K}$, or the negative change in potential energy $\mathbf{-\Delta U}$. With this logic, we would assume that the potential energy could just be written as $\mathbf{-\Delta U = \int_a^b F dx}$. This is CLOSE, but still INCORRECT.

warning

$\mathbf{-\Delta U = \int_a^b F dx}$ is NOT a valid equation for potential energy under the influence of an electric field.

The path through which the force is applied can be a line or a curve. Thus, a path integral is used to find the potential energy. Here, instead of linear displacemnt $\mathbf{x}$, the path is instead represented by the infinitesmal lines which create the path, represented by $\mathbf{s}$. Therefore, the correct path integral becomes $\mathbf{-\Delta U = \int_a^b {F}_{e} ds}$

Key Equation

Potential energy for an electric field is defined as $\mathbf{\Delta U = -q_{0}\int_a^b E ds}$, where: $\mathbf{E}$ is the electric field vector, or the symbol representing the field $q_{0}$ is the positive test charge which we placed in the field $\Delta U$ is the potential energy $\mathbf{a}$ marks the first point on the path $\mathbf{b}$ marks the second point on the path $\mathbf{s}$ marks the infinitesmal lines which make the path

If you divide the potential energy by the test charge, you get a value which depends on the source charge distribution. This value is called electric potential and only depends on the source charge distribution and has a value at every point of an electric field. The equation $\mathbf{V = \frac{\mathbf{\Delta U}}{q_{0}}}$ represents this concept. The change in electric potential is called potential difference and is represented by $\mathbf{\Delta V}$, which is $\mathbf{V_{b} - V_{a}}$. Through this, the eqution of the potential difference becomes $\mathbf{\Delta V = \frac{-q_{0} \int_a^b E \, ds}{q_{0}}}$ or $\mathbf{\Delta V = -\int_a^b E ds}$

9.1.2 - Potential Differences in a Uniform Electric Field

9.1.3 - Obtaining the Value of the Electric Field from Electric Potential