HelloooOoo World...

What is two bodies of different masses (1. Lighter mass, 2. Heavier mass) subjected to same force placed on a frictionless smooth surface. Does any mass accelerate in applying force? Acceleration is different but does every mass containing object accelerate??

Does always any massvie object accelerate in applying force? Or it behave as friction force as static force, limiting force, kinetic force. Pls clear....

@EmilioPisanty 39 so far and they are still coming :-)

Presumably they've just fixed some bug that wasn't awarding Steward badges?

is anyone getting a Steward badge everyday?

@EmilioPisanty so what changed?

Ok, so it's given for every thousand reviews instead of just once for the first thousand.

Hi @JohnRennie Sir.

Pls see my question. Object having mass, is there always acceleration (no matter how small depend on mass) any isolated object with only applied force.

@RewCie loool. Online class?

Help

I have reached the diffeological space section of Schreiber

I may be dying

@JohnRennie @Yashas it's an intentional design change. Steward is now multi-award, once per 1k tasks.

It seems to have caught up now. I got a total of 40 gold badges out of it :-)

@JohnRennie that's... a lot.

Congratulations!

whats a steward badge?

@JohnRennie Heavy is the crown

:-)

@satan29 physics.stackexchange.com/help/badges/107/steward

Is the formula for Coulomb’s law here incorrect, specifically the vector inside the charge density?

Following the notation of Griffith’s book.

no, that seems to be correct...

Anyone here knows where to find the answers of "Thomas Calculus" exercises? The back of that book has answers to only odd numbered problems -__-

@Sid yes buddy.... XD ROFL!

@schn It is correct, yes

a quick sanity check is putting $\rho(\mathbf{r}-\mathbf{r}') = q\delta(\mathbf{r}-\mathbf{r}')$

@Slereah @satan29 how come it is correct? In Griffith’s edition 3 it is:

@schn they're equivalent

note that it's $\hat{r}$ there

so it's normalised

@NiharKarve But in Griffith’s it is only $\mathbf r’$ in $\rho()$, or?

"It shall be unlawful, except as provided in section 2121 of this title, for any person, inside or outside of the United States, to knowingly participate in the development of, manufacture, produce, transfer, acquire, receive, possess, import, export, or use, or possess and threaten to use, any atomic weapon."

Well there goes my weekend

@schn How does he define $r'$?

He uses like $r'$, $r$, $\hat{r}$ and $\tau'$

It's a bit of a mess

He also uses the script r thing, which I only hated writing second to $\xi$ (and can't even seem to figure out the tex for)

@DanielUnderwood It's an ugly hack - it's just a graphic and not a font symbol, Griffiths has a link at the bottom of his webpage for his code for that

$\mathbf r’$ is where the source is at, since that is what is being integrated with respect to. $\mathbf r$ is the field point, and the displacement vector is the difference between them...if I’m not mistaken

In light of that, @Slereah, would you say my original image of the formula for the E field was incorrect? Intuitively, you’d want to know the charge density at the source, not anywhere else, right?

I'm in the process of renaming and categorising all of my PDFs

it's gonna take me an hour for the next two weeks

such is the price for order

let chaos reign

I would have

but finding a relevant paper is a nightmare when they all have descriptive names like "9712.0923" and "Smith"

ah shoot

my laptop has this weird setting where it doesn't put the quotes until you press it twice

@NiharKarve just curious, is your last name pronounced "car - way" ?

Also, I had a question from Electromagnetism

My professor discussed boundary-value problems, where we use the Laplace equation to find the resistancce

@satan29 कर्वे

Aha, thought so :-)

now, $\nabla ^2 V =0$ inside the region of the conductor is justifiid by the use of the continuity equation and by ohms law:

$\nabla . \vec{J}=-\dfrac{\partial \rho}{\partial t}=0$ , and $\vec{J} = \sigma \vec{E}$

@Slereah what's the main new thing that you gained that you absolutely couldn't do without in the sections before diffeological spaces of the nlab stuff

but then by gauss law $\nabla .\vec{E}= \rho / \epsilon_{0} $ which gives $\nabla ^2 V = \rho / \epsilon_{0}$ since $\vec{E}=-\nabla V$

which means $\rho$ is zero everywhere? thats just wrong

so gauss law fails????

whats going on here

@satan29 why is it wrong? there is usually no net charge in conductors

huh

theres definitely a charge density if current is flowing...

@satan29 you're not supposed to resolve the conductor into its individual charged ions and electrons here, you're just averaging their charge over the body of the conductor, and that average charge is zero

the current flowing doesn't change that there's no net charge since as much charge flows into the conductor as is leaving it that way