electric potential
Electric potential
Electric potential is defined as electric potential energy per unit of electric charge. More precisely, electric potential is the amount of work needed to move a test charge from a reference point to a specific point in a static electric field. The test charge used is small enough that disturbance to the field is unnoticeable, and its motion across the field is supposed to proceed with negligible acceleration. Electric potential should not be confused with electric potential difference.
Definition
Electric potential, \[V\], is the electric potential energy per unit electric charge, i.e. \[V=\frac{U}{q}\]. Since it is shown that \[U=-\int_{C}q\mathbf{E}\cdot d\mathbf{s}=\frac{1}{4\pi\epsilon_{0}}\frac{Qq}{r}\], thus \[V=-\int_{C}\mathbf{E}\cdot d\mathbf{s}=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r}\].
Electric field as the gradient of potential
The electrical potential difference \[V\] measured over a path \[C\] is given by \[V=-\int_{C}\mathbf{E}(\mathbf{r})\cdot d\mathbf{s}\] where \[\mathbf{E}(\mathbf{r})\] is the electric field intensity at each point \[\mathbf{r}\] along \[C\]. We begin by identifying the contribution of an infinitesimal length of the integral in \[V\], \[dV=-\mathbf{E}(\mathbf{r})\cdot d\mathbf{s}\]. Expanding \[d\mathbf{s}\] with Cartesian coordinates, we get \[d\mathbf{s}=\hat{\mathbf{x}}dx+\hat{\mathbf{y}}dy+\hat{\mathbf{z}}dz\]. We also note that for any scalar function of position, in this case \[V(\mathbf{r})\], it is true that \[dV=\frac{\partial V}{\partial x}dx+\frac{\partial V}{\partial y}dy+\frac{\partial V}{\partial z}dz\]. Substituting \[dx=d\mathbf{s}\cdot \hat{\mathbf{x}}\] and so on gives us \[dV=\frac{\partial V}{\partial x}(d\mathbf{s}\cdot \hat{\mathbf{x}})+\frac{\partial V}{\partial y}(d\mathbf{s}\cdot \hat{\mathbf{y}})+\frac{\partial V}{\partial z}(d\mathbf{s}\cdot \hat{\mathbf{z}})\]. We can rearrange this equation into \[dV=\left( \frac{\partial V}{\partial x}\hat{\mathbf{x}}+\frac{\partial V}{\partial y}\hat{\mathbf{y}}+\frac{\partial V}{\partial z}\hat{\mathbf{z}} \right)\cdot d\mathbf{s}\].
Comparing both \[dV=-\mathbf{E}(\mathbf{r})\cdot d\mathbf{s}\] and \[dV=\left( \frac{\partial V}{\partial x}\hat{\mathbf{x}}+\frac{\partial V}{\partial y}\hat{\mathbf{y}}+\frac{\partial V}{\partial z}\hat{\mathbf{z}} \right)\cdot d\mathbf{s}\] gives us \[\mathbf{E}(\mathbf{r})=-\left( \frac{\partial V}{\partial x}\hat{\mathbf{x}}+\frac{\partial V}{\partial y}\hat{\mathbf{y}}+\frac{\partial V}{\partial z}\hat{\mathbf{z}} \right)\]. Then, by definition of gradient we can shorten this equation to \[\mathbf{E}(\mathbf{r})=-\nabla V\].
With this fancy new equation, we can utilize the gradient theorem to calculate \[\int_{C}\mathbf{E}\cdot d\mathbf{s}=-\int_{C}\nabla V\cdot d\mathbf{s}=-(V(\mathbf{b})-V(\mathbf{a}))\], assuming that the path \[C\] is a curve with endpoints \[\mathbf{a}\] and \[\mathbf{b}\].