Problem Solving Strategies

Lllt

04:22

@JohnRennie @satan29 Suppose a neutral conducting sphere and Q charge are placed at a center to center seperation r , what will be the potential due to the induced charge on sphere at it's pole

The answer by calculations is negative but shouldn't it be zero?

John Rennie

@Lllt Inside the sphere the field is zero. Yes?

Lllt

Yes @JohnRennie sir

John Rennie

And the field inside the sphere is the sum of:
1. the field due to the point charge
2. the field due to the charge induced on the sphere

Lllt

YES

The second one is zero at the center

We can find due to first one at center

Adding gives net potential

$\frac{kQ}{r}$

John Rennie

Yes. The net potential moving from the front of the sphere to the back of the sphere is zero because the field is zero.

So subtract off the change in potential caused by the point charge and you have the change in potential caused by the field of the sphere.

Lllt

04:34

Yes I did and got the answer

$$\frac{kQ}{r}-\frac{kQ}{\sqrt{r^2+R^2}}$$

This shows that -ve and +ve induced charge's distribution are not same

John Rennie

Correct :-)

Lllt

But why does this happens that the induced charge distribution is not same

John Rennie

Because the field that the sphere is sitting in is not uniform.

If you put the sphere in a uniform electric field then you would get a symmetrical charge distribution.

Lllt

Yes I got that it's clear Thankyou :-)

Negative induced charge will increase more rapidly than the positive charge , am I right?

John Rennie

:-)

caramalizedTomato

05:09

@JohnRennie Could you help me with a question?

John Rennie

@caramalizedTomato Yes, of course. What's the question?

caramalizedTomato

I have posted it, above Lllt's message

John Rennie

11 hours ago, by caramalizedTomato

11 hours ago, by caramalizedTomato

Why is it that we ignore the component of velocity while writing the eqn for the element dx?

caramalizedTomato

Yep, btw how do you do that? I mean reply with my message visible?

John Rennie

caramalizedTomato

05:14

39 secs ago, by John Rennie

John Rennie

:-)

caramalizedTomato

Alright now I know it

@JohnRennie So what don't understand is why the induced EMF is indipendant of the angle of the small element dx?

John Rennie

Anyhow, I'm not sure what you mean by:

> Why is it that we ignore the component of velocity while writing the eqn for the element dx?

Adil Mohammed

05:32

@Lllt Is the charge inside the sphere or outside?

caramalizedTomato

@JohnRennie Sorry to keep you waiting

John Rennie

Ah, OK, so for each length d𝓁 of the wire you consider the components of the velocity parallel to and normal to the wire.

caramalizedTomato

So what I am trying to ask is, is the EMF induced in the small part dx of the parabolic rod dependant of the sine of the angle?

@JohnRennie Why parallel component?

John Rennie

The two components of the velocity produce EMFs that are normal to each other. The component normal to the wire produces an EMF that runs along the length of the wire, and this component will end up being summed along the wire to produce the total potential difference between the ends of the wire.

The component parallel to the wire produces an EMF that is directed sideways to the wire i.e. it creates a potential difference between the two edges of the wire. This component does not contribute to the potential difference between the ends of the wire.

OK so far?

caramalizedTomato

Yes

John Rennie

05:40

> is the EMF induced in the small part dx of the parabolic rod dependant of the sine of the angle?

So the EMF measured along the length of the wire is dependent on the sine of the angle.

Lllt

@AdilMohammed Outside

John Rennie

But I guess you're asking something slightly different because you're considering the distance dx measured along the y axis not the distance d𝓁 measured along the length of the wire. Yes?

(I can draw a diagram if it isn't clear what I mean)

caramalizedTomato

Actually I was trying to do the latter but just ended up creating the wrong diagram

Along the length of the wire indeed

John Rennie

The potential difference dV along the length of the wire d𝓁 does depend on sin θ, but if you consider the distance dx along the vertical axis then the sin θ is going to cancel out and disappear.

caramalizedTomato

Is it axis specific? I mean if I chose the x axis it should give me the same result ie. the cancelling of the sin$\theta$

John Rennie

05:48

The problem is that along the x axis there is no change in the potential.

The change in the potential happens normal to the velocity.

caramalizedTomato

@JohnRennie How so? $dl$ does have a component in $dx$

John Rennie

Consider this diagram ...

satan 29

@JohnRennie B,D ?

John Rennie

I've drawn in two more rods shown by the red and blue lines.

caramalizedTomato

Ok

John Rennie

05:58

All three rods have the same velocity V₀ î

The potential difference between the ends of the blue rod is zero. Yes?

satan 29

@caramalizedTomato the equation for motional EMF is $\mathbf{v} \times \mathbf{B} .\mathbf{dl}$

caramalizedTomato

@satan29 Also C

@JohnRennie Yes

John Rennie

And if we combine the red and blue rods they begin and end at the same place as the curved rod.

So the PD across the red rod has to be the same as the PD across the curved rod.

caramalizedTomato

@satan29 yes

@JohnRennie That's what the integral would tell us right?

satan 29

vxB will be in y direction, so we will end up with $\hat{y}.\mathbf{dl}$ which means we care about the y component only

John Rennie

06:02

Part C is the purple rod, and the same argument applies.

So as Satan29 says, only the displacement in the y direction matters.

caramalizedTomato

@JohnRennie So any rod of any structure with the same vertical distance b/w the ends will be equivalent to this case

John Rennie

Yes

satan 29

exactly

caramalizedTomato

Got it
Thnx @JohnRennie @satan29

satan 29

$B_{0}V_{0} \int_{y1}^{y2} (1+ (y/L)^{\beta} dy$ is the equation you get finally. as long as y1 and y2 are same, the EMF of the rod will be the same no matter the shape

John Rennie

06:08

@caramalizedTomato :-)

caramalizedTomato

@satan29 Yeah that seems plausible now

@JohnRennie one more thing, a current carrying wire experiences a force in presence of magnetic field

John Rennie

Yes

caramalizedTomato

38 mins ago, by John Rennie

The component parallel to the wire produces an EMF that is directed sideways to the wire i.e. it creates a potential difference between the two edges of the wire. This component does not contribute to the potential difference between the ends of the wire.

Is this what causes it?

John Rennie

All these forces originate from the Lorentz force.

If you have a wire carrying a current then the conduction electrons in that wire are moving at the drift velocity. Yes?

caramalizedTomato

@JohnRennie Yeah I mean not the EMF itself but the force from which this EMF along the edges is generated

John Rennie

06:21

So in a stationary wire carrying a current we have electrons moving at the drift velocity v relative to the field B and we get a Lorentz force on each electron F = ev x B.

These forces on the individual electrons sum up to give the total force on the wire.

Integrate this force and you get the potential difference.

caramalizedTomato

Aha, I understand it now thank you sir
=)

John Rennie

:-)

Lllt

08:42

@JohnRennie Sir there?

John Rennie

@Lllt Hi :-)

Lllt

Hello sir

Suppose we have two conductors

One having charge Q1 and potential V1 and other having charge Q2 and potential V2

Suppose we coonect them with a conducting wire

John Rennie

OK ... ?

Lllt

Now why will charge flow from higher potential to lower potential, I know it's a fundamental property but how does that happen here

Is there some electric field from higher potential conductor to lower potential conductor

John Rennie

If the two conductors are at different potentials then when we connect them with a wire there will be a potential gradient along the wire i.e. the potential must change continuously as we move along the wire from one conductor to the other. Yes?

Lllt

08:46

One minute

Yes

John Rennie

So if we calculate dV/dx, where x is the distance along the wire, then dV/dx ≠ 0.

Lllt

Yes

and this dv/dx gives electric field along the wire

John Rennie

Exactly :-)

Are you thinking that because the wire is a conductor the field inside it should be zero?

Lllt

Yes

John Rennie

The field is zero inside a conductor at equilibrium

i.e. when no currents are flowing.

And when we first connect the wire obviously the system is not at equilibrium and a current is going to flow along the wire.

Lllt

08:50

when we say potential of a conductor , we mean potential of every point on it?

??

John Rennie

Let's make an analogy with a container of gas. At equilibrium the pressure is the same everywhere in the container and the gas is stationary. Yes?

Lllt

Yes

John Rennie

So in this this case the pressure on the container is easy to define because the pressure is the same everywhere inside the container.

But now suppose we have a partition down the middle of the container and we have different pressures either side of the container.

Lllt

Yes

John Rennie

Then at time zero we remove the partition.

For the short time it takes the gas to equilibrate the pressure is different in different parts of the container. Yes?

Lllt

08:55

ok so at time dt we cannot tell exact pressure near the places where the partition was there

John Rennie

Exactly.

Lllt

@JohnRennie Yes

John Rennie

In this case it doesn't make sense to ask "what is the pressure of the container" because the pressure is not constant in the container.

Lllt

Yes

John Rennie

I guess you could ask "what is the average pressure", but this is a different question.

Lllt

08:56

Yes

No I wanted to ask what does the term potential of conductor mean, Does it mean the potential of every point on it?

John Rennie

And this applies to a conductor. If it is at equilibrium, i.e. the current is zero, then the potential is the same everywhere in the conductor and it makes sense to ask what the potential of the conductor is.

> Does it mean the potential of every point on it?

At equilibrium the potential is the same at every point. Yes?

Lllt

Yes

John Rennie

So the potential of the conductor is the same as the potential everywhere in the conductor.

Because every point in the conductor has the same potential.

There is only one "potential" for the whole conductor, and we take that as the potential of the conductor.

Lllt

@JohnRennie Why it will change continuously

John Rennie

If the potential changes discontinuously that means E = dV/dx = ∞. Yes?

Lllt

09:01

Yes

John Rennie

But you cannot have an infinite field in any material. It would exert an infinite force on the electrons in the material and that force would tear the material to pieces.

Lllt

Yes

John Rennie

The field must be finite, and therefore dV/dx cannot be infinite.

Lllt

and therefore dv/dx must have some value everywhere

John Rennie

Yes

Lllt

09:05

So when we connect two conductors and we connect them the charge flows from higher potential to lower potential and to find electric field we can do -dv/dx

John Rennie

Yes

Lllt

Clear Sir :-)

John Rennie

:-)

Lllt

@JohnRennie Sir will there be no charge on wire at electrostatic condidtions

John Rennie

When the two conductors and the wire have settled down to an equilibrium state the potential will be the same everywhere, including the wire.

This means the potential of the wire is non-zero. Yes?

Lllt

09:15

Yes

John Rennie

Imagine we now disconnect the wire from the two spheres (or whatever the two conductors are). We can do this without changing anything because if the potential is the same everywhere we can disconnect the wire without changing the potential. OK so far?

Lllt

Yes

John Rennie

So, we now have a piece of wire at some potential V. But before we started, i.e. before we connected the wire, our piece of wire had a potential of zero. So the potential of the wire has now changed. Yes?

Lllt

Yes

John Rennie

And that change can only have happened because the charge on the wire changed from zero to some non-zero value when we connected it. So our connecting wire must be charged after we have connected it.

Lllt

09:20

yes

John Rennie

9 mins ago, by Lllt

@JohnRennie Sir will there be no charge on wire at electrostatic condidtions

So the answer to your question is "no".

Lllt

Yes clear , there will be some charge on the wire

John Rennie

:-)

Robin Singh

10:01

@RaMathuzen what do u mean by "we can only use P_o*(l-h)"?

RaMathuzen

10:59

@RobinSingh What I meant was, When you submerge the capillary tube in water because of surface tension of water there will be a capillary rise, let's say a height H (H < l) Then you seal the open end of the capillary So now the product "PV" for the air trapped is $P_{0} \cdot A (l - H) $

where A is the Cross-sectional Area and $P_{0} $ is the atmospheric pressure. Then now you submerge the capillary a distance "x" in water such that the level of water is same inside and outside the capillary So the product "PV" for air trapped now is P' \cdot A (l-x) $ , Right ?

@RobinSingh

Robin Singh

@RaMathuzen I reckon that first they sealed the tube and then put it in water coz note that l=110mm but water actually rises to a height of 10.3m. So if what u say is correct then the water would just come out of the upper open end of the tube in the first case

RaMathuzen

@RobinSingh Yeah

Thanks

Robin Singh

@RaMathuzen Is that a "yea thanks I understood" yeah or a "bad worded question and i am not really satisfied" yeah? xD

RaMathuzen

11:14

The former

I did calculate H now after you said

Ali

21:33

Suppose there is a mixture of gases kept in a container of total volume V which is pressed by a piston which exerts constant pressure of P. There is one gas of n1 moles occupying partial volume V1 and exerting partial pressure of P1.

So, we know that P×V=N(total moles of all gases)×R×T

We also know that P1×V1=n1RT

We also know that P1×V =n1RT(Dalton's law)

And also P1× V =n1RT(Amagat's law of partial volume ).

Which of my equations can't be applied here, if any ? If all the equations are true, they give very stupid results.

@JohnRennie or @satan29 can you answer this ?

or anyone who wants to,please do

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Problem Solving Strategies