Remarks
- Remember: SI unit of energy is Joule (1 J).
- When the force does a positive work, the potential energy decreases.
- Only a difference of potential energy is really meaningful. We can always add up a constant. To give an absolute value, potential energy must be defined relative to some reference point where $U=0$. In circuits, this is called ground
- When working with atomic systems, it is useful to define another unit: a electron volt (1 eV); $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{J}=e^-\times 1V$
#Reinterpreting the electric potential
The electric potential can then be interpreted as a potential energy per unit charge
$$ V=\frac{U}{q_0} $$ and a voltage as a work per unit charge $$ \frac{W_{a \rightarrow b}}{q_{0}}=-\frac{\Delta U}{q_{0}}=-\left(\frac{U_{b}}{q_{0}}-\frac{U_{a}}{q_{0}}\right)=-\left(V_{b}-V_{a}\right)=V_{a}-V_{b} $$And we find the following equivalence of units
$1 \mathrm{V}=1$ volt $=1 \mathrm{J} / \mathrm{C}=1$ joule/coulomb
#Energy in Uniform Fields
$$ U=q_{0} E y $$![[Pasted image 20240306144313.png]]
#Electric Potential Energy of Two Point Charges
By setting U to zero when two point charges $q$ and $q_0$ are infinitely far apart $r=\infty$, we have
$$ U=\frac{1}{4 \pi \epsilon_{0}} \frac{q q_{0}}{r} $$- The potential energy U given is a shared property of the two charges. In general, its a property of the charge configuration.
- The equation also holds if the test charge $q_0$ is outside a spherically symmetric charge distribution with net charge q; This is because Gauss’s law states us that the electric field outside such a distribution is the same as if all of its charge q were concentrated at its center.
- The energy of a collection of charges is the algebraic sum (energy is additive!) of each pair of charges.
Polarization