# Taha Azzaoui - 2017.07.04

### The Electric Force

We understand by means of experimentation that two point charges separated by a distance r exhibit a force proportional to the inverse of r2. More specifically, according to Coulomb’s law, we can express this electric force as
$$\boldsymbol{F}_{e} = \frac{1}{4\pi\epsilon_{0}} \frac{q_1 q_2}{r^2} \hat{r} \tag{1}$$
Where ϵ0 is the permittivity of free space and qi is the charge contained by the ith point charge. Note that Fe is negative when the charges differ in sign and positive otherwise. This coincides with the intuitive notion that opposites attract.

### The Conservative Nature of the Electric Force

It is important to note that the electric force is a conservative one. This is a consequence of the fact that the work done by the electric force along a path that starts and ends at the same point is identically zero. To see this, recall the equation for the work done by a force
$$W = \oint_l \boldsymbol{F} \cdot d\boldsymbol{l} \tag{2}$$
If we substitute the electric force from (1) into (2) and express dl in spherical coordinates, we get the following
$$\begin{eqnarray} W &=& \oint_s \frac{1}{4\pi\epsilon_{0}} \frac{q_1 q_2}{r^2} \hat{r} \cdot (d\boldsymbol{r}\hat{r} + rd\boldsymbol{\theta}\hat{\theta} + rsin\theta d\boldsymbol{\phi}\hat{\phi}) \nonumber \\ &=& \frac{q_1q_2}{4\pi\epsilon_{0}} \int \frac{1}{r^2} dr \nonumber \tag{3} \end{eqnarray}$$
Finally, we evaluate the resulting integral along a closed path that starts and ends at the same point p.
$$\frac{q_1q_2}{4\pi\epsilon_{0}} \int_p^p \frac{1}{r^2} dr = \frac{-q_1q_2}{2\pi\epsilon_{0}} \frac{1}{r^3} \Bigr\|_{p}^{p} = 0 \tag{4}$$
Therefore, we have shown that the work done by the electric force along a path that starts and ends at the same point is identically 0, which implies that the electric force is conservative.

### The Electric Field

With a non-contact force such as the electric force, it can be difficult to visualize the effect of a point charge (or a collection of them) on its surroundings. Consequently, we introduce the concept of an electric field, a vector quantity which we define to be the electric force per unit charge at every point in space. In other words, if a probe charge q0 experiences a force Fe at some point in space (x, y, z), then the electric field at that location is
$$\boldsymbol{E}(x, y, z) = \frac{\boldsymbol{F_e}}{q_0} \tag{5}$$
Note that we now have a new way of expressing the electric force
$$\boldsymbol{F_e}= \boldsymbol{E} q_0 \tag{6}$$

### Electric Potential Energy

Another way of visualizing the effects of charge on its surroundings is via the scalar potential. It follows from the conservative nature of the electric force that there must exist some potential energy function U associated with it. Furthermore, by the work-energy theorem, the net decrease in potential energy must be equivalent to the work done by the electric force.
$$dU = - \boldsymbol{F_e} \cdot d\boldsymbol{s} \tag{7}$$
Similar to the way we defined the electric field, we now introduce a new quantity, the electric potential difference, which we define to be the change in potential energy per unit charge.
$$dV = \frac{dU}{q_0} \tag{8}$$
One may be suspicious of the significance of equations (7) and (8). These equations are useful in that they allow us to describe properties of the electric field without dealing with the underlying vector calculus. Therefore, we now have two ways of describing the electric field- the first is using the definition of the electric field itself (equation 5) and the second is using the electric potential difference (equation 8). We now describe the relationship between these two methods and demonstrate how to go back and forth between them. We start by rewriting equation (7) using equation (6)
$$dU = - \boldsymbol{E} q_0 \cdot d\boldsymbol{s} \tag{9}$$
Next, we substitute this equation into (8)
$$dV = -\boldsymbol{E} \cdot d\boldsymbol{s} \tag{10}$$
We can express this equation in component form
$$\begin{eqnarray} dV &=& -(E_x\hat{i} + E_y\hat{j} + E_z\hat{k}) \cdot (dx\hat{i} + dy\hat{j} + dz\hat{k}) \nonumber \\ &=& -(E_x dx + E_y dy + E_zdz) \nonumber \tag{11} \end{eqnarray}$$
This relationship implies
$$\begin{eqnarray} \boldsymbol{E} &=& -(\frac{\partial{V}}{\partial{x}}\hat{i} + \frac{\partial{V}}{\partial{y}}\hat{j} + \frac{\partial{V}}{\partial{z}}\hat{k}) \nonumber \\ &=& -(\frac{\partial}{\partial{x}}\hat{i} + \frac{\partial}{\partial{y}}\hat{j} + \frac{\partial}{\partial{z}}\hat{k})V \nonumber \\ &=& - \nabla V \nonumber \tag{12} \end{eqnarray}$$
In other words, the electric field has magnitude equal to the rate of change of the electric potential and points in the direction of maximum potential decrease.