- #1

ynbo0813

- 31

- 4

*If*I understand correctly, Gauss’ Law is (roughly) derived as follows:

**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)

*know*I must have made a misstep, misunderstanding, or false assumption somewhere. I’m just trying to figure out what it is.