AP Physics C-E&M: Calculating q_encl

[darksyesider]

Homework Statement

http://media.collegeboard.com/digitalServices/pdf/ap/ap14_frq_Physics_C-E-M.pdf

#3 c ii)

Homework Equations

The Attempt at a Solution

$\oint E\cdot dA = \dfrac{q_{encl}}{\epsilon_0}$

To find q_encl would this be correct;

$q_{encl} = \int \dfrac{-\beta}{r^2}\cdot e^{-r/\alpha}\cdot (4\pi r^2)dr$

then integrating that, and substituting it into the expression for q_{encl}.$E = \dfrac{q_{encl} }{\epsilon_0 \cdot 4\pi r^2}$

 

[Simon Bridge]

The calculation do not appear to relate to E&M 3 c ii.
You are given the charge density already. What is the relationship between that and the charge enclosed?

 

[darksyesider]

Well the charge density is not the actual charge (q enclosed), so i integrated the expression of:
density * surface area *dr to get a thin spherical shell of charge which accumulates. This would be q enclosed.

Is this correct?

 

[Simon Bridge]

The q-enclosed is the charge enclosed inside the gaussian surface.
It sounds like you have the right approach to work it out.

Note: your second line is the expression for q-enclosed.
You then say "integrate that and substitute into the expression for q-enclosed" ... i.e. into itself.
This is what threw me off.

Presumably you mean - substitute into into another expression to find E.

 

[Redbelly98]

The Attempt at a Solution

$\oint E\cdot dA = \dfrac{q_{encl}}{\epsilon_0}$

To find q_encl would this be correct;

$q_{encl} = \int \dfrac{-\beta}{r^2}\cdot e^{-r/\alpha}\cdot (4\pi r^2)dr$

Partially correct. Reread the very first sentence of the problem. Besides the equation for the electron charge density, what else does the model consist of?

then integrating that, and substituting it into the expression for q_{encl}.


$E = \dfrac{q_{encl} }{\epsilon_0 \cdot 4\pi r^2}$

Yes, that is the idea.