- #1
ynbo0813
- 31
- 4
If I understand correctly, Gauss’ Law is (roughly) derived as follows:
Part A
Now, Gauss’ Law is applied to cylinders as follows:
Part B
Part C
Part A
- Electric Flux = EA
- E = q / (∈4πr^2)
- A of the surface of a sphere is 4πr^2
- They cancel out and therefore EA =q/∈
Now, Gauss’ Law is applied to cylinders as follows:
Part B
- A of a cylinder is 2πrL
- EA of a cylinder = E2πrL
- EA is also = q/∈ (from 4 in Part A)
- q/∈ = E2πrL (from 2 and 3 in Part B)
- Solve for E = q/∈2πrL
- q = λL
- E = λ / ∈2πr
Part C
- Electric Flux = EA
- E = q / (∈4πr^2)
- A of the surface of a sphere is
- EA = q2πrL / (∈4πr^2)
- EA = qL / ∈2r (or λL^2 / ∈2r)