PHYS 208 Lecture 4

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Electric Field

{\vec {E}}({\vec {r}})={\frac {kq}{r^{2}}}\,{\hat {r}}

{\vec {E}}_{\mathrm {total} }=\sum {\vec {E}}_{i}=\sum _{i=0}^{n}{\frac {kq_{i}}{{r_{i}}^{2}}}\,{\hat {r}}_{i}

{\vec {E}}_{\mathrm {total} }=\int \mathrm {d} {\vec {E}}=\int _{\mathrm {distribution} }{\frac {k\mathrm {d} q}{r^{2}}}{\hat {r}}

Uniformly distributed charge $Q$ along a line of length $l$ :

$\mathrm {d} q=\lambda \mathrm {d} x={\frac {Q}{l}}$

Therefore, the total field at a point $P$ is:

${\vec {E}}=\int {\frac {kQ\mathrm {d} x}{lr^{2}}}$

Imaginary lines to graphically represent the Electric field in the space around a charge

Properties and Corollaries

points away from a positive charge and points toward a negative charge
- Start at positive charges and end at negative charges
Direction aways tangent to the electric field at that point
strength is proportional to number of lines in the area
- parallel lines of equal density represent a uniform electric field
Lines will never cross

The field lines between two similarly charged particles will behave asymptotically at the plane between them.

the rate of transfer of fluid, particles, or energy across a given surface ^[1]

Take the integral over a surface...?

\Phi =\oint {\vec {E}}\cdot \mathrm {d} {\vec {A}}

$\mathrm {d} {\vec {A}}=\mathrm {d} A{\hat {n}}$ is perpendicular to the surface
Units = Nm²/C
Open surface Electric field lines cross through a single plane
Closed surface enters through plane and exits through plane on opposite side of object
Proportional to the number of field lines going into a shape (influx; counted negative) and the number of field lines coming out of a shape (outflux; counted positive)
No net charge inside surface: flux = 0

For any surface with any net charge (any distribution) inside, the flux is

\Phi =EA=\oint {\vec {E}}\cdot \mathrm {d} {\vec {A}}={\frac {Q_{\mathrm {net} }}{\epsilon _{0}}}