Physics
Kevin MehallElectric Flux
8.02 Lecture 3d\Phi = \vec{E} \cdot \hat{n} dA = \vec{E} dA cos \theta
Closed Surface
Sphere:
\Phi = 4 \pi R^2 E = \frac{Q}{\epsilon_0}
Gauss's Law:
\Phi = \oint \vec{E}. \vec{dA} = \frac{\sum Q_{inside}}{\epsilon_0}
Sphere of radius R with charge Q uniformly distributed at surface:
Symmetry Argument 1:
all points at a given radius r must have equal field
Symmetry Argument 2:
field must point radially outward or inward, so angle is 0 or 180
At points r\lt R:
4 \pi R^2 E = \frac{Q_{inside}}{\epsilon_0}
Q_{inside} is 0, so E=0
no electric field inside charged sphere
At points r>R:
4 \pi R^2 E = \frac{Q_{inside}}{\epsilon_0}
Q_{inside} = Q, so E = \frac{Q}{4\pi r^2 \epsilon_0}
from the outside, a charged sphere acts the same as if all charge were at center
Flat Horizontal Plane of infinite size:
Charge density
= \sigma = \frac{Q}{A} (\frac{C}{m^2})
Find electric field at point height d
Use for Gauss surface:
Cylinder with top and bottom parallel to plane, with top surface distance d above plane and bottom surface d below plane
No flux goes through sides of cylinder
E = \frac{\sigma}{2 \epsilon_0}
Electric field does not fall off with distance
Two oppositely charged planes separated by distance d
Top plate has charge +\sigma, bottom -\sigma
Field above and below are 0, inside E = \frac{\sigma}{\epsilon_0}
