PHYS 208 Lecture 6

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Conductors

Electric fields inside conductors is always 0; charge always resides on the surface of the conductor

Electric Potential

Review of Mechanics

Work of a force: $\underset{a}{\overset{b}{\int }}\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{l}$

For conservative forces, $W=-(U_b-U_a)=-\mathrm{\Delta }U$

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Electric forces are conservative forces!

Application of Work to Coulomb's law:

$W=\underset{a}{\overset{b}{\int }}\frac{kQq}{r^2}\hat{r}\cdot \mathrm{d}\overrightarrow{r}=-(\frac{kQq}{r_b}-\frac{kQq}{r_a})$

Therefore, the potential energy of an electric charge is

$U(r)=\frac{kQq}{r}+C$

Note: potential of multiple charges is the sum of potential energies between all possible pair combinations

$U=\sum _{ij}k\frac{q_i{q}_j}{r_{ij}}$

Because electricity is much stronger than gravity, the constant of integration $C$ actually matters now.

Conservation of energy also applies:

$W=\int ({\overrightarrow{F}}_E+{\overrightarrow{F}}_{\mathrm{a}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{t}})\cdot \mathrm{d}\overrightarrow{l}=0$

Example

Moving a -1nc charge that is 1m away from a 5nc charge to a point 5m away:

$W=-[U(5m)-U(1m)]=-k{q}_1{q}_2[\frac{1}{5m}-\frac{1}{1m}]=-36\times 10^{-9}\mathrm{J}$

Work done my moving agent must be equal and opposite: 36 × 10⁻⁹ J (positive work — when the force and direction of movement are in the same direction)

Electric Potential for a point charge

Measured in J/C = volts [V]

$V(r)=\frac{kq}{r}+C$

Based on the units, dividing work by the charge results in the integral of the electric field:

$V=\frac{W}{q}=\underset{a}{\overset{b}{\int }}\frac{\overrightarrow{F}}{q}\cdot \mathrm{d}\overrightarrow{l}=\underset{a}{\overset{b}{\int }}\overrightarrow{E}\cdot \mathrm{d}l$

It is also important to note that the work divided by the charge is the negation of the change in electric potential:

$-\mathrm{\Delta }V=\frac{W}{q}⟶\mathrm{\Delta }V=-\underset{a}{\overset{b}{\int }}\overrightarrow{E}\cdot \mathrm{d}\overrightarrow{l}$

In order to get rid of the constant, we can use infinity as a reference point where $V=0$:

$V(r)=V(r)-V(\mathrm{\infty })=\frac{kq}{r}$

For a continuum (surface or solid) of point charges:

$V(r)=\int \frac{kdq}{|\overrightarrow{r}-\overrightarrow{r}{}^{\prime }|}$