Ch.5 Potential and Field

Relationship between Potential and Field

Potential and field are two different ways of describing how source charges alter the space around them.

Because $qV\equiv\Delta U=-W(\text{i}\rightarrow \text{f})=-\int_\text{i}^\text{f}\vec{F}\cdot d\vec{s}$ we can divide by $q$ and using the fact that $\vec{F}=q\vec{E}$ , we see that $V=-\int_\text{i}^\text{f}\vec{E}\cdot d\vec{s}$
Note the negative sign indicates that field decreases along the field direction
The reverse implies
$E_s=-\frac{dV}{ds}$
This shows the field along the displacement $\Delta s$ , so is useful when a coordinate axis can be determined so that the field perpendicular is $0$
Note that this is related to the $F_s=-\frac{dU}{ds}$ equation by a factor of $q$

$\vec{E}_{||}$ tangential to equipotential is $0$ .
$\vec{E}_{\perp}$ perpendicular to equipotential points in the decreasing direction
So in general,
$\vec{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}=-\left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right)=-\nabla V$

Kirchoff's Loop Law

sum of all potential differences in a closed loop or path is zero
$\Delta V_\text{loop}=\sum_i(\Delta V)_i=0$

Electrostatic Equilibrium

Since $\vec{E}=0$ inside a conductor in electrostatic equilibrium, $\Delta V=0$ , i.e. the potential is the same across the whole conductor

Properties

all excess charge is on the surface
surface is equipotential
electric field inside is $0$
the interior is equipotential
exterior electric field is perpendicular to the surface
surface charge density and electric field strength are largest at sharp corners

equipotential surface close to electrode roughly matches shape of electrode, and "morph" going from one electrode to another

Create Electric Potential

potential is created simply by separating positive and negative charge
$\Delta V=V_\text{pos}-V_\text{neg}=-\int_\text{neg}^\text{pos}E_sds$
Since electric force tries to bring positive and negative charge together, a nonelectrical process is required to separate charges

Van de Graaff generator moves a plastic or leather charged using a corona discharge to move charge to the positive spherical electrode
- electric motor running the belt does work to lift the charges, since electric force pushes it down
- potential is created between sphere and its surroundings
batteries use chemical reactions
- charge-escalator model: charge escalator lifts positive charge from negative to positive terminal (energy supplied from chemical reactions)
- work done per charge is emf: $\epsilon=W_\text{chem}/q$
- terminal voltage: charge separation creates potential difference $\Delta V_\text{bat}$ $Δ V_{bat}$ , where the ideal battery has $\Delta V_\text{bat}=\epsilon$ $Δ V_{bat} = ϵ$
  - when current runs through, $\Delta V_\text{bat}<\epsilon$ , but in most cases is negligible

Batteries in series produce $\Delta V_\text{series}=\Delta V_1+\Delta V_2+\cdots$

Capacitance

This $\Delta V$ is plausibly proportional to the charges $\pm Q$ that create the voltage difference, so set $Q=C\Delta V_C$ , and solve
$C=\frac{Q}{\Delta V_C}$
This is the capacitance with units of Farads $1\text{ F}=1\text{ C/V}$ , where $\Delta V_C$ is capacitor voltage

Parallel Plate Capacitor

Recall that the potential difference across a parallel-plate capacitor with distance $d$ and electric field $E$ is $\Delta V_C=Ed$ , and $E=Q/\epsilon_0 A$ from Gauss's Law. So,
$Q=\frac{\epsilon_0 A}{d}\Delta V_C\implies C=\frac{\epsilon_0 A}{d}$
The capacitance depends entirely on the geometry of the electrodes.

They can be charged by connecting to a battery. The $\Delta V_C$ steadily increases as the charge escalator transfers charge, until it reaches $\Delta V_\text{bat}$ (the repulsive force equals the battery escalator's force; electrostatic equilibrium)

They can be combined in parallel capacitors (top to top; bottom to bottom) or series capacitors (end to end in a row), represented in a circuit diagram by two paralle lines $\rightarrow$ have an equivalent capacitance
In parallel capacitors, the top two and bottom two electrodoes have the same potential as the top/bottom of the battery. So, we have $Q_1=C_1\Delta V_C$ and $Q_2=C_2\Delta V_C$ , but the battery moved a total of $Q_1+Q_2$ charge, so the capacitance is equivalent to havign $Q_1+Q_2=C_\text{eq}\Delta V_C\implies C_\text{eq}=C_1+C_2$
In general,
$C_\text{eq}=C_1+C_2+\cdots\space\text{parallel capacitors}$
In series, the wire between the capacitors is isolated, so have a net $0$ charge. Thus, the battery moves the same $Q$ charge to the top. The potential difference is the sum of the differneces $\Delta V_C=\Delta V_1+\Delta V_2$ , and using the fact that $\Delta V_1=Q/C_1$ and $\Delta V_2=Q/C_2$ , we have $1/C_\text{eq}=\Delta V_C/Q=1/C_1+1/C_2$
In generall,
$\frac{1}{C_\text{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots\space\text{series capacitors}$

Energy in a Capacitor

The potential energy change from "lifting" the positive charges is
$dU=dq\Delta V=\frac{qdq}{C}$
So the total potential energy is
$U_C=\frac{1}{C}\int_0^Q qdq=\frac{Q^2}{2C}=\frac{1}{2}C(\Delta V_C)^2$
The voltage of a capacitor $\Delta V_C=Ed$ and the capacitance $C=\epsilon_0 A/d$ can be substituted to get various other forms
$U_c=\frac{1}{2}C(\Delta V_C)^2=\frac{\epsilon_0 A}{2d}(Ed)^2=\frac{\epsilon_0}{2}(Ad)E^2$
The energy density $u_E$ of the electric field is the energy divided by the volume
$u_E=\frac{U_C}{Ad}=\frac{\epsilon_0}{2}E^2$
with units $\text{J}/\text{m}^3$

Dielectrics

Suppose a capacitor separated by a vacuum a distance of $d$ is charged to voltage $(\Delta V_C)_0$ then disconnected, giving each plate a $\pm Q_0$ charge where $Q_0=C_0(\Delta V_C)_0$ .
If the vacuum is replaced by another insulating material, a dielectric, the voltage will decrease, but the charge remains the same, so the capacitance increases.

So, if the plates are filled with an insulator, the capacitance increases.
The insulator in the electric field gets polarized, represented by two sheets of charge with surface charge densities $\pm\eta_\text{induced}$ : induced electric field
$\vec{E}_\text{induced}=\frac{\eta_\text{induced}}{\epsilon_0},\space\text{inside the insulator from positive to negative}$
This field points opposite the capacitor's field, weakening the overall field inside the capacitor.
The dielectric constant $\kappa$ is
$\kappa\equiv\frac{E_0}{E}$
where $E_0$ is the field when the capacitor has a vacuum inside and $E$ is the final field. Since $E$ is weaker, $\kappa\ge1$
$\kappa$ is the factor by which the field is weakened; consider that $\epsilon_0$ gets multiplied by the constant

$\kappa$ is a property of matrials, like specific heat or density, and more easily polarized materials have a higher $\kappa$

The voltage is divided by $\kappa$ and the capacitance is multiplied by $\kappa$

$\eta_\text{induced}=\eta_0\left(1-\frac{1}{\kappa}\right)$
where $\eta_0$ is the surface charge density on the capacitor plates.

The dielectric strength is the maximum electric field the material can handle before breakdown- production of a spark.