ATAR Notes: Forum

Uni Stuff => Science => Faculties => Physics => Topic started by: /0 on July 04, 2010, 06:46:05 am

Title: Electrostatics - direction of integration
Post by: /0 on July 04, 2010, 06:46:05 am: If you have a line of charge with charge density $\lambda=\frac{dq}{dl}$ and you want to find the electric field at a perpendicular distance z from the midpoint, you get

$dE = \frac{1}{4\pi\epsilon_0}\frac{\lambda}{r^2}dl$

Then you integrate $dE$ from one end of the line of charge to the other. (e.g. $\int_{-L}^L ... dl$ )

Obviously if you reverse the integral terminals you get the negative of your original answer, but physically, why should reversing integral terminals even matter?

After all, the physical interpretation of the integral is just summing up the little $dq$ s over the line, what does it matter which direction you do it in? And importantly, how do you know which is the correct direction to sum up the $dq$ s?
Title: Re: Electrostatics - direction of integration
Post by: mark_alec on July 04, 2010, 12:41:19 pm: For an (infinite) line of charge, you should get the result (using a cylindrical gaussian surface) that $E(r) = \frac{\lambda}{2 \pi\epsilon_0 r}\hat{r}$

I'm not sure what you are trying to integrate along, if you do it along an equidistant path from the line charge then the integral should be zero (as the potential is constant).
Title: Re: Electrostatics - direction of integration
Post by: kamil9876 on July 24, 2010, 11:00:07 pm: To know why, you have to be careful with your definition of an integral from a BIGGER to SMALLER terminal. Take a look at this sum. so the dq's are those differences, you need these to be positive, in other words $x_{i+1}>x_i$ and so you are going forwards.

In either case, you integrate towards the way which the axis is pointing.
Title: Re: Electrostatics - direction of integration
Post by: Cthulhu on August 12, 2010, 12:15:32 am: Just because I cbf making a new electrodynamics thread:

can somone check this answer for me?

Quote
Find the electric field inside a sphere which carries a charge density proportional to the distance from the origin,
$\rho = k r^3$ , for some constant $k$ .
So...... We can use

$\oint_S \boldsymbol{E} \cdot \mathrm{d}\boldsymbol{a} = \frac{1}{\epsilon_0}Q_{enc}$
The charge inside the sphere is
$Q_{enc} = \int \rho \; \mathrm{d}\tau = \int_0^{2\pi}\mathrm{d}\phi \; \int_0^\pi \sin\theta \; \mathrm{d}\theta \; \int_0^r kr'^3 \mathrm{d}r'$
Where $r'$ is just an integrating variable.
$\therefore Q_{enc} = k \times 2 \pi \times 2 \times \frac{r^4}{4} = k\pi r^4$

And the electric field is the same in all points on the sphere so we can say
$\int \boldsymbol{E} \cdot \mathrm{d}\boldsymbol{a} = \int |E|da = |E|\int da = |E|4\pi r^2$
and so
$|E|4\pi r^2 = \frac{1}{\epsilon_0}k\pi r^4$
$\therefore E = \frac{1}{4 \epsilon_0}k r^2$
or
$\boldsymbol{E} = \frac{1}{4\epsilon_0} k r^2 \boldsymbol{\hat{r}}$

amirite?

edit: fixed the last line.
Title: Re: Electrostatics - direction of integration
Post by: /0 on August 12, 2010, 03:46:38 am: When integrating over a sphere I think $dV = r^2\sin\theta\,d\theta\,d\phi\,dr$

(Funny, I did a very similar problem for homework except that $\mathbf{E} = kr^3\mathbf{\hat{r}}$ and we had to find the enclosed charge.)
Title: Re: Electrostatics - direction of integration
Post by: Cthulhu on August 12, 2010, 11:02:15 am: Quote from: /0 on August 12, 2010, 03:46:38 am
When integrating over a sphere I think $dV = r^2\sin\theta\,d\theta\,d\phi\,dr$

(Funny, I did a very similar problem for homework except that $\mathbf{E} = kr^3\mathbf{\hat{r}}$ and we had to find the enclosed charge.)

haha. Funny how i forgot $r^2$ I'd already answered the question and was just typing it out again from memory.

I guess it should be

$E = \frac{1}{6 \epsilon_0} kr^4$