@JohnRennie do u follow Brian Greene?

@JackRod I read his first book, The Elegant Universe, and I thought it was very good.

@JohnRennie Good Morning

@Mayank hi :-)

What is the logic behind deducing Coulomb's law by Gauss law?

@JohnRennie

@Mayank The two are effectively equivalent i.e. different ways if stating the same theory.

Ok

@JohnRennie What is the relation between charge density and electric field?

@Mayank electromagnetism is described by four equations known as Maxwell's equations, and it's the first of these that relates the electric field to the charge density. The trouble is it uses maths called vector calculus that you probably haven't learned yet.

:(

On the up side, think of all the fun you'll have learning this stuff in the next few years.

Which level?

You'd probably learn vector calculus when you start university, though it's not that hard and all the JEE students I know have enough knowledge of maths to learn it if they wan,ted to.

I wanted to know it for this

You don't need to know vector calculus to understand that.

He is just saying that if you pack more charge into a given volume then it generates a stronger field, which is kind of obvious really.

:)

@JohnRennie any inputs here? ...have been on PSE for over a month but still no strong answer

@AnindyaPrithvi Nothing immediately springs to mind ...

@JohnRennie damped oscillations? :P

I think it depends on how you define the potential, especially since it isn't conservative for the loop ina changing magnetic field.

You could connect two points with a high resistance wire and ask what current flows in the wire, then define the potential difference as the current times the resistance of the wire.

@JohnRennie what's the potential difference between two very close points?

considering the V=iR definition

I guess you could consider a single electron moving in a circle and consider the Lorentz force on it. Then the potential change would be the force times the distance.

I told about two very close points because there can be two values of distance....one is approx zero, second is 2piR

@JohnRennie A single electron will have uncertaininty, how can I decide the distance-path it follows

i.e. you go round the loop n times.

@JohnRennie that's the fun part....how would a potentiometer behave?

I am surprised i could not find any experiments on this

nor any proposal to research

It's been nearly 12 months since I first had this doubt

Any device for measuring the potential is basically a high resistance wire joining the two points. A perfect galvanometer would have an infinite resistance while real galvanometers have a finite resistance.

So join the two points with a resistance R and calculate the current in it. Then take the limit of R going to infinity (assuming that limit is well defined).

@JohnRennie what will be the current...can it be predicted...even if i assume a real galvanometer/voltmeter?

Yes that should be straightforward calculation.

How

I don't know offhand how to do the calculation, but it can obviously be done since in a real experiment the current would have a well defined value.

@JohnRennie how are you so certain about the well defined value

Believe me, I have been trying to predict every possible way to carry out the experiment in my head

but all lead to a multivalues output

Well it's an experiment you could do. And assuming the experiment is reproducible there must be a well defined value for the current.

Do college labs have a set-up to see the TVMF?

@JohnRennie measuring the state of a qubit is possible...but it is not necessarily the same always

If I set the basis vectors as 0,1....and measure along the Y basis

I would get + or - depending on what the "nature" wants me to see

Right but this is a purely classical system. You don't expect to see quantum effects on a macroscopic system.

@JohnRennie I would not have expected if I received atleast a rudimentary mathematical proof

There's lorentz here...wont be surprised to see QM as well

If instead of a conducting-ring, it were n loops of insulated wire...neglicting the edge effects, I am certain to get a single value...but not in this case

@AnindyaPrithvi suppose you draw a diagram like this:

And suppose there is a changing magnetic field $\dot B$ normal to the page

Call the upper loop $1$ and the lower loop $2$, and the resistance of the circular wire $R_a$ and the resistance of the radial wire $R_b$.

We'll eventually want to take the limit $R_b/R_a \to \infty$

We know the EMF round the loops because it's just $E = -A\dot B$, where $A$ is the area of the loop, so we can use Kirchoff's laws to calculate the three currents.

And once we know $I_3$ we can calculate the potential difference between the two points where the radial lines intersect the circle.

@JohnRennie If one doesn't want to use kirchoff law...going from the center of the circle to the edge of the circle...the potential is zero....and why wouldnt it be

@JohnRennie Secondly, it might not be a great idea to place it at the center....And...why would you use kirchoff law even stating that potential is not defined?

sorry for the late response..I muted my device and was giving a mock test

Hmm, I get the current in the radial lines always equal to zero. Though I ran through the algebra quickly an may well have made a mistake.

So...how to proceed..I do not have access to any labs or fund to set up a high end experiment

Just do the calculation.

Assuming I didn't make a mistake $I_3 = 0$ so the potential you would measure between any two points on the circle is zero.

@JohnRennie just like the case depicted in the IRODOV problem in that question

so we went from current to no current....regardless of the resistance of the voltmeter

Yes

I just used Kirchoff to calculate $I_3$ and got $I_3 = 0$

@JohnRennie funny question part two: if I stack such loops in a cylindrical sense and put a charge of magnitude +q at the centre of the the cylinder, the rings would all be equipotential with a distinct value in each loop. so, if I connect all the loops, the delocalized charges in the plane of the centre would move up destroying the electric field at the centre?

@JohnRennie Hi

@Mayank hi :-)

What would be the work if there is acceleration?

@JohnRennie

When it says there is no acceleration it means the kinetic energy is not changing.

The total energy is the PE plus the KE, and the change in the total energy has to be equal to the work done. Yes?

Yes

If there is no change in the KE then the change in the total energy is equal to the change in the PE. So in this case the work done is equal to the change in the PE, which is what your book is saying.

If there is acceleration that means the KE changes (because if there is acceleration the velocity changes)

In that case we would have to write W = ΔPE + ΔKE

So Can we say that W=Fdcosø=∆PE+∆KE?

@JohnRennie

Yes

Looks like I have a misconception

@Mayank In what way?

My book says

"When a coolie is carrying some load on his head moves on a horizontal platform . Work done by the coolie is zero"

How it can say that Work done by coolie is zero?

What if he is accelerating?

It's assuming the coolie is moving at constant speed.

Why don't they write it?

"When a coolie is carrying some load on his head moves on a horizontal platform at constant speed. Work done by the coolie is zero"

@Mayank don't know - ask the author of the book :-)

If he is accelerating

FdcosØ=0?

No

If he is accelerating in a horizontal direction then there must be a horizontal force acting. Yes?

Yes

Oh I got it now.

:-)

The total force is the vector sum of the vertical gravitational force and the horizontal force

Hmmm

The displacement is horizontal, and the work is W = F . d, where the . is the dot product.

If we write F = Fv + Fh, where Fv is the (vertical) gravitational force and Fh is the horizontal force causing the displacement then we get:

W = (Fv + Fh) . d = Fv.d + Fh.d

And Fv.d = Fv d cos 90° = 0

But Fh.d = Fh d cos 0° = Fh d

So the work is W = Fh d

Yes ... ?

Do you think this is a good way to teach work?

When students are understanding it first time

That's a rather basic introduction. I would use vectors and say that work is given by the dot product W = F.x

I guess that means you need to know about vectors and maybe students who are just starting don't know this.

But I think vectors are simple enough that even young students could learn about them.

That's 9th grade book

What age is 9th grade?

14-15

I can't remember when I first learned about vectors. It was a long time ago :-)

@JohnRennie hi sir :-) How are you ?

@ronakjain hi :-)

@JohnRennie sir , can you help me in organic chemistry ?

If he is accelerating, where is the horizontal force acting?

@JohnRennie

The man's velocity is a vector. Yes?

Unless he is jumping up and down the direction of the velocity is horizontal.

Yes

And if he is accelerating then the velocity is changing. If we assume he accelerates in the same direction as he is already moving then the direction of his velocity doesn't change and only the magnitude changes. Yes?

Yes

So the direction of the acceleration is the same as the direction of the velocity.

I've drawn the acceleration in blue.

OK so far?

Wait.

Some network problem

I was thinking

OK ...

Force by his legs on ground and then reaction force

@JohnRennie

Is acceleration involved in all forces?

Remember that Newton's second law says F = ma

So if there is an acceleration there must always be a force associated with it.

Is vice versa possible?

Or looking at it the other way round, if we apply a force F there will always be an acceleration given by a = F/m

And the mass m is just a number, not a vector, so the direction of the acceleration vector is always the same as the direction of the force vector.

@JohnRennie Then how electrostatic force is acting without producing acceleration?

The electrostatic force does produce an acceleration. What makes you think that it doesn't?

So how it is possible to displace a test charge without acceleration?

@Mayank You mean when your book talks about no acceleration?

Yes

I think I am asking a silly question

If you put a positive test charge q near another positive charge Q and just let go of it then the test charge is repelled by the electrostatic force and accelerates away. So what happens is the PE of the charge is converted to KE. OK so far?

No

@JohnRennie You said KE is constant.

@JohnRennie Ok

I'm working up to explaining that. What I'm saying here is that no external forces are being applied so the test charge feels only the force F = kQq/r² repelling it from the charge Q so it accelerates away and its KE increases.

Okay

The increase in the KE is the same as the decrease in the PE so the total energy remains constant.

Ok

Decrease in KE?

What your book is saying is suppose you hold the charge q and you move it slowly at a constant velocity so the charge cannot accelerate.

@Mayank oops!

Decrease in PE of course.

Anyway, as I said above, imaging you are holding the charge so it can't move reely and you control the motion by moving your hand holding the charge.

It means you grab a hold of it with your hand.

Ok

Now if you are holding the charge that means the charge is exerting a force on your hand. Yes?

Yes. Reaction force?

Yes

So if you now move your hand outwards that force does work on your hand.

Hmmm

That work would be W = ∫ F dr

Yes

If you slowly move the charge out to infinity then the PE changes from kQq/r to zero, so the PE has decreased. But the KE didn't increase because you were holding the charge and stopping it from accelerating. So where did the PE go?

And the answer is that it went into the work done on your hand.

So we are exerting external force

Yes

Electrical Potential energy is confusing topic.@JohnRennie

I think potential energy and work are confusing in all areas. It's something you only get used to with experience.

@JohnRennie Have you ever seen the word 'Electric Mechanical Energy'' in any textbook?

No

It's my request that if anyone knows the answer, please confirm it. Because I'm already very confused regarding the topic.

@SarGe Electric field is zero in the body of the conductor...i.e. (ii)