@JohnRennie please see my question in the main site. I have examination so I can't stay here for long.

@DivyankaChaudhari what have you done here?

@sammygerbil thanks

@DivyankaChaudhari You made that assertion that flux entering charge q2 cannot be at an angle more than alpha by seeing the diagram right?

Or is there another way?

How did you come up with that?

@DivyankaChaudhari But how many exactly?

@DivyankaChaudhari See acc to you it should be:

Flux leaving from solid angle associated with angle $\beta$ = flux terminating in solid angle associated with $\alpha$

My question is:

Why can't it be:

Flux entering from solid angle associated with $\alpha +\text{ x degrees}$ = flux terminating in solid angle associated with $\beta$ , where x is some other angle

@DivyankaChaudhari edited

like why not an angle greater than alpha

@DivyankaChaudhari why not

@DivyankaChaudhari Nah, confused about it.

Okay, thanks.

@DivyankaChaudhari sis, I have just pinged him.

@DivyankaChaudhari ok

@user187604 Morning :-)

@JohnRennie hi,sir

@Akash.B Morning :-)

@JohnRennie sir please hang out

putting some some effort in researching a question

@JohnRennie sir , why sun appears to be coming from below earth during the sunrise?

isn't the sun should appear to come from the side

@JohnRennie sir you didn't understood it properly

I will give you a link

@JohnRennie sir I will be available for next 3 hours

@DivyankaChaudhari nice question

@PrathamArya Hi :-)

hello

You were asking about the potential across two capacitors in series?

yes

What was the problem?

I was thinking if two capacitors have the same area, distance between plates and same charge then they should have same capacitance

but when connected in series there arise different potentials

so why capacitance changed when it is independent of Q and V

@JohnRennie sir ,you there?

That's what the potentials will look like for two identical capacitors in series.

V/2 ?

The potential difference across each of the capacitors will be V/2 i.e. the same for both

i am unable to understand

@PrathamArya Suppose you take a voltmeter and connect the negative terminal to the anode of the battery, then the voltages written on the right will be what the meter shows at that point.

@JohnRennie 2 basic questions... when can I ask?

Ask now ...

@pratham potential difference across both caps is same

can I say the potential of a capacitor as work done in moving a positive charge from negative end to the positive plate of that capacitor?

@JohnRennie If gravity was in upwards direction, and the pendulum was kept in the usual way (with the bob facing downwards), would SHM be possible?

I don't think it would be.

@Abcd the bob would be at a position of unstable equilibrium i.e. if you displaced it the tiniest amount it would accelerate away from its starting point.

Just like with normal gravity if the bob was at the top.

@PrathamArya yes, it's the work done moving a unit charge i.e. one coulomb

@JohnRennie =No SHM right?

@Abcd I'm not sure what you're asking ...

@Abcd have you tried making force diagrams?

Imagine normal gravity with the bob starting at the top. It would swing right down to the bottom then back up to the top again.

@PrathamArya yes, I can see no restoring force. I just needed verification.

(I assume this is a rod not a string - with a string the bob would just fall straight down)

@JohnRennie rod with a bob?

Let me show you the actual question then...

Consider normal gravity i.e. pointing down. The situation you describe is the same as normal gravity with the bob starting at the top instead of the bottom. Yes?

yes

@JohnRennie Thanks sir ,i got the point

Comparing with $T= 2\pi \sqrt{\dfrac{l}{g}}$

Here $l= a$

$g= qE/m$

$\implies T = 2 \pi \sqrt{\dfrac{am}{qe}}$

Which is the right answer.

But I am wondering whether it should be -qE or +qE

I feel it should be -qE because only then situation would be perfectly analogous to gravitational situation

@JohnRennie That's my actual problem^

@JohnRennie Oh, I had kept it vertially

@Abcd whatever. The orientation doesn't matter.

Did you understand my method @JohnRennie ?

But you're quite correct that it is just like gravity, except that gravity always pulls in the same direction while in this case the force on the $+q$ is right and te force on the $-q$ is left.

yes exactly

But I think you've got the right answer by luck!

@JohnRennie so $+q$ should be clamped right?

For a start the dipole pivots about the centre, so the length is $a/2$.

@Abcd no, neither charge is clamped

@JohnRennie charges are at the ends of the rod

and one end is clamped

The rod pivots about the point shown by the arrow

@JohnRennie Why has he written "rod is clamped at an end"

@JohnRennie I am listening.

Ah, OK, sorry I didn't read the question properly. Yes, it does say the rod is clamped at one end.

OK, in that case it is exactly like a pendulum but with $qE$ replacing $mg$ as the force.

Am I right?

So the answer should be: $T = 2 \pi \sqrt{\dfrac{ma}{|-qE|}} $

No, the force is always $F = qE$, though of course one or both of $q$ and $E$ can be negative.

Yes

But my point is ... which charge is the force acting on

-q or +q

Ah, I see what you're asking.

Note that $E$ is vertically upwards

so if Its $+q$

the situation will be like upward gravity

It doesn't matter which end you clamp.

Please give me a minute, I'll be back

@JohnRennie Please see^

Call the charge $Q$, where $|Q| = |q|$ and $Q$ can be positive or negative

Ok

Then the force is $F=QE$

Yes

But if q is -ve

F = -QE = downwards = just like gravity

So just put $QE$ in your equation

If q is +ve,

F= QE = upwards = like "upwards gravity"

OK I guess I assumed the question means the dipole is aligned with the field

because that is the position of stable equilibrium

That's how I've drawn my diagram.

Ok

If the dipole is aligned opposite to the field then that's a position of untable equilibrium and it won't move with SHM.

Instead it will flip right round so it's aligned with the field.

@JohnRennie I had drawn the same stable equilibrium in vertical direction,

@JohnRennie My book proves that there's no charge inside a charged conductor using Gauss' law and choosing Gaussian surface "just below" the surface of the conductor

@JohnRennie Using the same method if we choose the Gaussian surface tracing the conductor then wouldn't it prove something weird: there's no charge inside a charged conductor

I can't see how that proof works.

It is a proof that there is no field inside a charged conductor

But that only works because it is assumed the charge is all at the surface of the conductor

@SmarthBansal Oh, duh. Yes, got it. Thanks.

@JohnRennie Why do we use Gauss' Law only for infinite distributions?

infinite distributions?

Gauss' law works for any charge distribution ...

It just relates the total charge inside the surface to the net field through the surface

@JohnRennie Even when the charge distribution is a function?

Yes.

Oh please give me a second

All that matters is the total charge inside the surface

ignore that "wont ...axis of the ring" part

So linear charge density is $\lambda = \lambda_o \cos \phi $

How can we use Gauss' Law for this problem @JohnRennie ?

Gauss' law isn't very use here

If you use a surface that completely encloses the ring then the total flux through the surface must be zero.

However that doesn't mean the field is zero every. It is zero overall because it's positive over one half of our Gaussian surface and negative over the other half, and the positive and negative bits add up to zero.

@DivyankaChaudhari re angles of electric field lines from charge +Q to smaller charge -q : all field lines ending on -q come from +Q. Other lines from +Q go to infinity. Charge +Q "emits" more field lines. Far from both charges the field looks like that of a +ve point charge (+Q-q). No field lines come from infinity and end on -q.

@JohnRennie Can we use it for a simple ring with +ve charge Q uniformly distributed on it? to calculate the field on its axis?

I can't think of a way to use Gauss' law for that ...

@Abcd I'll agree with @JohnRennie. This might be helpful

@SmarthBansal good find! :-) @Abcd that link explains in detail why Gauss' law isn't useful in that case.

@DivyankaChaudhari I think the gaussian surface must be closed. Not the distribution of charge.

@JohnRennie may I ask if you pinned the blog post from SO in the hbar, sir?

@skull No it wasn't me.

Thank you.

Sir.

@DivyankaChaudhari Sorry I misunderstood your question. Prathyush Poduval's answer in another question develops a formula which does answer your question. I note your handwritten solution at 12:01 AM. See "How to find the lines of force from electric field?" (physics.stackexchange.com/q/335044)

@JohnRennie could you please help me out in the question above?

@GaurangTandon What is your difficulty here?

@sammygerbil sorry, I forgot to copy-paste my problem in here

i don't understand what they mean by "equilibrium". I read this - en.wikipedia.org/wiki/Transient_equilibrium - but it doesn't seem to be what they're looking for.

this is my problem ---^

@GaurangTandon Th232 decays to Ra228 (not 238 as given in the question) which decays to Ac228 etc. Equilibrium means that the number of decays of Th232 per second equals the number of decays of Ra228 per second. ie The activity is the same. Then the amount of Ra232 is not changing.

@sammygerbil oh I see, thanks!

@GaurangTandon This situation is described as Secular Equilibrium (en.wikipedia.org/wiki/Secular_equilibrium) rather than Transient Equilibrium (as in your wiki link).

@JohnRennie sir are you there?

@sammygerbil oh that's complete, thank you!

@sammy thanks a lot, really helpful!

@JohnRennie please see the question!

Somebody see the question.

@user187604 What do you think? What is your difficulty?