Ch.4 Electric Potential

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Electric Potential Energy

Recall $W=\vec{F}\cdot\Delta\vec{r}$ and $\Delta U=-W$
Consider a parallel plate capacitor with a charge inside. Define the coordinates so that $s=0$ at the negative end and $s=d$ at the positive end, so that the field points in the $-s$ direction, like the gravitational field.
If a proton inside is moved from $s_i$ to $s_f$ along the electric field, then $W_\text{elec}=F\Delta r\cos0^\circ=qE(s_i-s_f)$. Thus, the electric potential energy changes by
$\Delta U_\text{elec}=-W_\text{elec}=qEs_f-qEs_i$
Note: electric field force is an internal force, so $U=-W$

In general, the electric potential energy of a charge in a parallel-plate capacitor is
$U_\text{elec}=qEs$
where $s$ is measured from the negative plate

Potential energy of a two-point charge (or two charged sphere) system
$U=\frac{kq_1q_2}{r}$

With negatively charged particles, $U<0$

The turning point is where $U=E_\text{mech}$
If $E_\text{mech}=0$, oppositely-charged particles can reach infinite separation at infinitesimally slow speeds

Note that internal work causes negative change in potential energy, but external work causes changes in total system mechanical energy

electric force is a conservative force; i.e. work done is independent on the path taken

total potential energy of multiple particles is sum over all pairs

Potential Energy of a Dipole

We have $dW=\tau d\phi$ and $\tau=-pE\sin\phi$, so total work is
$W=\int_{\phi_i}^{\phi_f} dW=\int_{\phi_i}^{\phi_f}\tau d\phi=\int_{\phi_i}^{\phi_f}-pE\sin\phi d\phi=pE\cos\phi_f-pE\cos\phi_i$
Thus $\Delta U_\text{elec}=-W_\text{elec}=pE\cos\phi_i-pE\cos\phi_f$, which yields
$U_\text{elec}=-\vec{p}\cdot\vec{E}$

Electric Potential

We define the electric potential or just "potential" $V$ as
$V=\frac{U_{q\text{ with sources}}}{q}$ Thus
$U_\text{elec}=qV$
The unit is Joules per Coulomb, or the volt

In an electric potential $V$, if a positive charge moves from a region of lower to higher potential, the potential energy must increase, therefore the velocity must decrease. Common, this is the potential difference, or voltage; thus particle decreases velocity as it moves through a positive potential difference

Parallel-Plate Capacitor

From $U=qEs$, we have $V_\text{cap}=Es$, where $s$ is the distance from the negative plate (increases closer to positive)
Assign $V_-=0$ and $V_+=Ed$ ($d$ is distance between plates), and we have the potential difference$\Delta V_C=Ed-0=Ed\implies E=\frac{\Delta V_C}{d}$
Thus, we can say $V_\text{cap}=\frac{s}{d}\Delta V_C$, showing the potential increases linearly with distance from the negative plate.

Graphical Representations of Capacitor Potential

• potential versus $s$
• 3-D equipotential surfaces graph
  • mathematical surfaces where each point on the surface has the ame potent
  • parallel to capacitor plates
  • equally spaced-out voltages
• 2-D contour map
  • essentially a edge view of the equipotential surfaces
• 3-D elevation graph
  • 3 dimensional view of potential vs $s$ graph, with a generalized "$yz$-graph"
• electric field lines point in the direction of decreasing potential and perpendicular to equipotential surfaces

Electric Potential of a Point Charge

outside a sphere of charge $Q$ and radius $R$, the potential is the same as that of a point charge at the center
$V=\frac{1}{4\pi\epsilon_0}\frac{Q}{r}$
On the surface, the voltage $V_0$ has $r=R$ and is "charged to a potential of $V_0$
$V_0=\frac{Q}{4\pi\epsilon_0R}$
Thus, $Q=4\pi\epsilon_0V_0R$, making the potential outside the sphere
$V=\frac{R}{r}V_0$

Multiple Charges

total potential is the sum of potentials due to each charge
integrating with a continuous charge distribution is easier because no need to account for components