Lesson 7 - Potential Energy and Voltage

1. It’s not “delta”

\[\vec{\nabla} = \frac{\partial}{\partial x} \mathbf{\hat{a}_x} + \frac{\partial}{\partial y} \mathbf{\hat{a}_y} + \frac{\partial}{\partial z} \mathbf{\hat{a}_z}\]

What are the names of this operator and its symbol?

operator	del
symbol	nabla

1.1. Divergence

\[\vec{\nabla} \bullet \vec{A} \quad\text{or}\quad \mathop{div} \vec{A}\]

Describe what it means that “the divergence of a vector field represents the presence of a source or sink of the vector \(\vec{A}(x, y, z)\) at each point.”

Nice try! Pretty sure you clicked before answering yourself ;)

Really do try to come up with a description NOT using “source/sink”. Then click to see a Dr. White version.

Take a vector field, like a particular electric field in space. You have enough experience to visually pick out the charges, perhaps, by seeing where the arrows point towards or away from a region.

Remember that the field lines are FAKE — they do not exist in any tangible way. The lines simply trace a particular path that a massless charge would travel if released at some location.

Given a vector field that you know nothing else about it besides its vector-value at each point in the region, the divergence of the field is like a searchlight that reveals sources or sinks of the field’s “mojo” (really, energy). For an electric field, these are positive and negative (point) charges. Divergence itself is the more general operation and can find other types of sources/sinks for other types of fields — consider: “gravitational field”.

1.2. Gradient

\[\vec{\nabla} \vec{A} \quad\text{or}\quad \mathop{grad} \vec{A}\]

It does look a little weird to have the vector arrow over the del for the gradient, but del really is a vector operator and it is useful to remind ourselves of this fact.

Cartesian

\[\nabla\vec{A} = \frac{\partial}{\partial x}\vec{A} \!\cdot\! \mathbf{\hat{a}_x} \; + \; \frac{\partial}{\partial y}\vec{A} \!\cdot\! \mathbf{\hat{a}_y} \; + \; \frac{\partial}{\partial z}\vec{A} \!\cdot\! \mathbf{\hat{a}_z}\]

Cylindrical

\[\nabla\vec{A} = \frac{\partial}{\partial \rho}\vec{A} \!\cdot\! \mathbf{\hat{a}_\rho} \; + \; \frac{1}{\rho} \frac{\partial}{\partial \phi}\vec{A} \!\cdot\! \mathbf{\hat{a}_\phi} \; + \; \frac{\partial}{\partial z }\vec{A} \!\cdot\! \mathbf{\hat{a}_z}\]

Spherical

\[\nabla\vec{A} = \frac{\partial}{\partial r }\vec{A} \!\cdot\! \mathbf{\hat{a}_r} \; + \; \frac{1}{r} \frac{\partial}{\partial \theta}\vec{A} \!\cdot\! \mathbf{\hat{a}_\theta} \; + \; \frac{1}{r\sin{\theta}} \frac{\partial}{\partial \phi }\vec{A} \!\cdot\! \mathbf{\hat{a}_\phi}\]

The order of the unit vectors is intentional here, it defines a right-handed coordinate system. Swapping any one pair ordering will yield a left-handed system, which is not the usual convention.

1.3. Laplacian

What happens when you take the divergence of the gradient of a scalar function: \(\mathop{div} (\mathop{grad} A) = \vec{\nabla} \bullet \vec{\nabla} A\) ? ^[1]

\[\begin{align} \vec{\nabla} \bullet \vec{\nabla} &= \left(\frac{\partial}{\partial x} \mathbf{\hat{a}_x} + \frac{\partial}{\partial y} \mathbf{\hat{a}_y} + \frac{\partial}{\partial z} \mathbf{\hat{a}_z}\right) \bullet \left(\frac{\partial}{\partial x} \mathbf{\hat{a}_x} + \frac{\partial}{\partial y} \mathbf{\hat{a}_y} + \frac{\partial}{\partial z} \mathbf{\hat{a}_z}\right) \\ % &= \frac{\partial^2}{\partial x^2} \mathbf{\hat{a}_x \bullet \hat{a}_x} + \frac{\partial^2}{\partial x \partial y} \mathbf{\hat{a}_x \bullet \hat{a}_y} + \frac{\partial^2}{\partial x \partial z} \mathbf{\hat{a}_x \bullet \hat{a}_z} + \cdots \\ \nonumber \\ \nabla^2 &= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \end{align}\]

Notice that the vector symbol above the nabla disappears, and notice how that it is appropriate to do so.

2. Potential energy

forces between charges
E-field to represent what-if placement of charge in the presence of others
Gauss’s Law > Coulomb’s Law (?!? fields > charges, fields before forces !?!)

2.1. Equation 7.5

\[\Delta W = - \int\limits_{-\infty}^{x_b} \frac{Q_1 Q_2}{4\pi\epsilon_0 \left(x - x_a\right)^2}\mathrm{d}x \Rightarrow \frac{Q_1 Q_2}{4\pi\epsilon_0 \left|x_b - x_a\right|} \label{eq-7.5}\]

See equation (5) at Integral Table (.com).
The absolute value simply captures the effect of changing the perspective from a to b, it’s the distance between them.

2.2. Work

\[\Delta W = V_1 \cdot Q_2 \quad \text{(Equation 7.7)}\label{eq-7.7}\]

Given equation \(\eqref{eq-7.7}\), what are the fundamental units for voltage?

\(V_1 = \Delta W / Q_2\), so joules per coulomb.

potential energy: The stored energy in the system that can be converted to other forms of energy.
potential: The quantity that exists at every point around (a (collection of) charge(s)) that will create potential energy if another (charge) is brought into the region.

\[W_e = \frac{\epsilon_0}{2} \int\limits_{\Delta v}^{x_b} E^2 \mathrm{d}v\]

Why/How did the \(\frac{1}{2}\) factor show up?

Consider the relationship between equation \(\eqref{eq-7.5}\) and the definition of “potential” and the collection of charges.

2.3. Challenge 07

Explain and show, in your own words, the path from book (Equation 7.16) to (Equation 7.17). Then how Gauss’s Law shows up in 7.18. Then how the math works to jump to the conclusion shown in (Equation 7.19).

3. Voltage

\[\Delta V_{ab} = \int\limits_a^b \vec{E} \bullet \mathrm{d}\vec{l}\]

Lesson’s video at 44:41

Example 7.4: Is the coordinate system drawn on the figure right or left-handed?

KVL!

\[\oint \vec{E} \bullet \mathrm{d}\vec{l} = 0\]

4. E-field and voltage

\[\vec{E} = - \vec{\nabla} V \quad \text{or} \quad \mathop{grad} V\]

5. Poisson’s equation

\[\nabla^2 V = \nabla \bullet \nabla \frac{-\rho_v}{\epsilon_0}\]

Does it matter the order of the dot product part that makes up the Laplacian operator?

6. Summary table

Print or bookmark that table!

Table 1. Epic summary table from Chapter 7
↓Need \| Have→	\(\mathbf{\rho_v}\)	\(\mathbf{\vec{E}}\)	\(\mathbf{V}\)
\(\mathbf{\rho_v}\)		\(\vec{\nabla}\bullet \vec{E} = \dfrac{\rho_v}{\epsilon_0}\)	\(\nabla^2 V = - \dfrac{\rho_v}{\epsilon_0}\)
\(\mathbf{\vec{E}}\)	\({\displaystyle \vec{E} = \frac{1}{4\pi\epsilon_0} \int_{\Delta v} \frac{\rho_v(r')}{R^2}\hat{a}_R\,\mathrm{d}v'}\)		\(\vec{E} = - \nabla V\)
\(\mathbf{V}\)	\({\displaystyle V(x, y, z) = \frac{1}{4\pi\epsilon_0} \int_{\Delta v} \frac{\rho_v(x', y', z')}{R} \,\mathrm{d}x' \mathrm{d}y' \mathrm{d}z'} \\ \text{ with } \;R\!=\!\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}\)	\({\displaystyle \Delta V_{ab} = \int_a^b \vec{E}\bullet \mathrm{d}\vec{l}}\)
\(\mathbf{W_e}\)	\({\displaystyle W_e = \frac{1}{2} \int_{\Delta v} \rho_v V \,\mathrm{d}v}\)	\({\displaystyle W_e = \frac{\epsilon_0}{2} \int_{\Delta v} E^2 \,\mathrm{d}v}\)	\({\displaystyle W_e = \frac{1}{2} \int_{\Delta v} \rho_v V \,\mathrm{d}v}\)

1. Yes, a scalar function. We took the gradient of the voltage, remember, which is a scalar function.