Problem Solving Strategies

user8718165

04:47

Good morning sir :-) @JohnRennie

yuvraj singh

Morning JR SIR

user8718165

Good morning Yuvraj :-) @yuvrajsingh

John Rennie

@yuvrajsingh @user8718165 morning :-)

user8718165

@JohnRennie sir are you drinking coffee?

John Rennie

I'm about to make my first coffee of the morning,but if you want to ask something go ahead.

user8718165

04:51

@JohnRennie sir, yesterday's question. But no, you first have your coffee... :-)

John Rennie

05:04

20 hours ago, by user8718165

@JohnRennie its written that the barrier voltage can't be measured because of unavailability of mobile charges...so how do the scientists measure the potential barrier of silicon and germanium?

@user8718165 that question?

user8718165

@JohnRennie yeah sir

John Rennie

What do you mean by the barrier voltage? The voltage developed across a PN junction?

user8718165

@JohnRennie yeah sir

John Rennie

That's easy. You just increase the voltage until the diode starts conducting and you measure what voltage is needed.

Are you familiar with the IV curves for forward biased diodes?

user8718165

@JohnRennie yeah sir...its written...but its by trial....is there no way of confirming :-)

@JohnRennie no sir...not yet...by today...

John Rennie

05:09

user8718165

@JohnRennie yeah sir...got it...

John Rennie

This is a typical IV curve for a diode. This is just one of many you'll find on the Internet.

user8718165

@JohnRennie okay

John Rennie

A voltage $V_d$ develops across the junction due to the formation of a depletion layer, so you need to apply a voltage equal to or greater than $V_d$ to get any current to flow across the junction.

user8718165

@JohnRennie sir, actually the current increases before threshold v :-/

John Rennie

05:11

This is the threshold voltage shown on the graph.

@user8718165 that's because electrons can tunnel across the barrier. The tunnelling probability increases as the barrier width decreases so we get a gradual increase in current as the barrier thins, rather than a sharp increase in current.

user8718165

@JohnRennie thanks sir

John Rennie

You can calculate the junction voltage by analysing the IV curve in detail, but basically it's the same as the threshold voltage.

The threshold voltage is an approximation to the junction voltage. If we wanted to measure it precisely we'd do a more accurate calculation. You'd use a model for the V curve in which the junction voltage was a parameter, then fit that model to the experimental data.

Although I confess I don't know exactly how that's done. I'm sure it's only a Google away though.

Nobody recognizeable

@JohnRennie hi.

John Rennie

@Nobodyrecognizeable hi

Nobody recognizeable

@JohnRennie ^^

John Rennie

05:26

Suppose the apogee of the satellite is at a distance $d$. You know that energy is conserved,so the total energy (PE + KE) of the satellite is constant. You can calculate the initial KE and PE because you're given the initial distance and velocity. OK so far?

Nobody recognizeable

@JohnRennie OK.

John Rennie

And you can calculate the PE at the apogee because it is just $U = -GMm/d$. So if you can find the KE at the apogee you can calculate $d$. Yes?

Nobody recognizeable

@JohnRennie yes.

John Rennie

And the way you calculate $d$ is to use conservation of angular momentum. At launch the velocity is tangent to the radius vector so the angular momentum is just $L = v_i^2/R$ where $v_i$ is the initial velocity.

At apogee the velocity is again normal to the radius vector $d$ so again $L=v_a^2/d$ where $v_a$ is the apogee velocity.

Nobody recognizeable

@JohnRennie OK.

John Rennie

05:33

So conservation of energy gives you one equation in $d$ and $v_a$, and conservation of angular momentum gives you a second equation in $d$and $v_a$. So you have two equations in two variables. Solve these to get $d$ and $v_a$.

What you are doing here is an application of the vis-viva equation i.e. using conservation of energy and angular momentum:

Vis-viva equation

In astrodynamics, the vis-viva equation, also referred to as orbital-energy-invariance law, is one of the equations that model the motion of orbiting bodies. It is the direct result of the principle of conservation of mechanical energy which applies when the only force acting on an object is its own weight. Vis viva (Latin for "living force") is a term from the history of mechanics, and it survives in this sole context. It represents the principle that the difference between the total work of the accelerating forces of a system and that of the retarding forces is equal to one half the vis viva...

Nobody recognizeable

@JohnRennie I get b.

user8718165

@JohnRennie got it now sir :-) Thank you very much for helping sir

John Rennie

@Nobodyrecognizeable does it match the answer key?

Nobody recognizeable

@JohnRennie yes.

John Rennie

@Nobodyrecognizeable BOOM! :-)

It was easier than you thought :-)

Nobody recognizeable

05:41

@JohnRennie from vis viva equation we are finding a right?

@JohnRennie certainly.

John Rennie

The vis-viva equation wouldn't answer your question. I mention it because it is derived using exactly the method you just used to answer the question.

Nobody recognizeable

@JohnRennie aren't we finding a?

John Rennie

No, because $a$ is the semimajor axis not the apogee distance.

If $d_a$ and $d_p$ are the apogee and perigee distances then $2a = d_a+ d_p$

Nobody recognizeable

@JohnRennie I get a ad R and d_a =2r so the problem is unrealistic.

John Rennie

You can calculate the perigee distance in exactly the same way you just calculated the apogee distance, though in this case the perigee distance would turn out to be below the surface of the Earth.

Oh,no wait, the perigee distance would just be $R$.

So you'd get $a = 1.5R$

No, sorry, wrong again.

Nobody recognizeable

05:49

@JohnRennie don't we take the sun stationary? So a=2r

John Rennie

The answer is the distance above the earth not the distance from the centre of the Earth. The apogee distance from the centre of the earth is $d_a = 3R$.

Nobody recognizeable

@JohnRennie won't that mean a circular orbit?

John Rennie

And the perigee distance is just $R$ so the distance between the apogee and perigee points is $d_a + d_p = 4R$. Yes?

Nobody recognizeable

@JohnRennie so indeed from surface of earth its 2r.

Ya I get $d_a=3r $ and $d_p=r$ when you measure from the centre of earth. Have to subtract r if you wanna get from surface.

@JohnRennie ^^

John Rennie

In the vis-viva equation the distance $r$ is the distance from the centre of the Earth.

You almost never use the distance from the surface in orbital mechanics.

Nobody recognizeable

05:55

@JohnRennie it's all clear now. Boom!

John Rennie

:-)

Nobody recognizeable

@JohnRennie Thanks for the help. Have a nice day. Bye

John Rennie

Bye :-)

yuvraj singh

08:00

@JohnRennie hi

John Rennie

@yuvrajsingh hi

yuvraj singh

I have a question on QM should I ask it here or in h bar @JohnRennie

John Rennie

h bar, unless it's a JEE question

user8718165

08:39

hello sir @JohnRennie

John Rennie

@user8718165 hi :-)

user8718165

@JohnRennie In an n type doped, when electrons leave the donor level for the conduction band, holes are made in donor level. Do those holes affect conductivity sir?

John Rennie

No, because the donor states don't form a band. The density of the donor states is very small compared to the silicon bulk, so they can't link up to form a band. Instead they form separate localised states where the holes are trapped and can't move.

The same applies to P doping. The electrons excited into the states just above the valence band are also trapped and can't move. That's why in a P type semiconductor only the holes conduct.

user8718165

@JohnRennie sir why can't those holes(or electrons in p type) move around sir? why are they trapped?

John Rennie

@user8718165 consider a gas - like air. It doesn't conduct electricity because the electrons are bound to the air molecules and can't move.

user8718165

08:47

@JohnRennie okay sir

John Rennie

Obviously the dopant molecules aren't a gas, but they resemble a gas in that they are widely separated and can't interact with each other.

user8718165

@JohnRennie got it sir

@JohnRennie thank you very much sir

John Rennie

So as in a gas the electron or hole remains bound to the dopant atom and can't move away.

user8718165

@JohnRennie thank you so much for helping sir

John Rennie

You're welcome :-)

user8718165

08:48

@JohnRennie yeah

user8718165

09:03

@JohnRennie sir will you please come here in the evening (afternoon)?

I will ask you some program questions if you're not tired :)

John Rennie

@user8718165 I'll almost certainly be around this afternoon. From about 16:00 UK time for a couple of hours.

user8718165

@JohnRennie yeah! okay sir :-)

yuvraj singh

12:06

@JohnRennie sir hi

Whenever you free please ping I have very small question. @JohnRennie

Jasmine

14:26

@JohnRennie Hello sir !

John Rennie

@Jasmine hi :-)

Aladdin

@JohnRennie hello

John Rennie

@Aladdin hi :-)

Aladdin

Are u freeish

John Rennie

I'm jut finishing lunch. Give me five minutes. I'll ping you when I' ready.

Jasmine

14:31

Aladdin

Sure

John Rennie

@Jasmine I'm not 100% sure what the question is asking, but I wonder if it's a conservation of angular momentum problem?

Physics freak

@JohnRennie Sir you had answered a similar question before can you please answer this one also. physics.stackexchange.com/questions/507422/…

John Rennie

@Jasmine Suppose the angular velocity of the disk after the string has become taut is $\omega$, then the velocity of the block is $v = R\omega$ so the angular momentum of the block is $L = mvR = mR^2\omega$.

Jasmine

@JohnRennie yes

John Rennie

14:39

The AM of the disk is $L = I\omega = \tfrac12MR^2\omega$, and because AM is conserved these two have to add up to the original AM of the disk $\tfrac12MR^2\omega_0$.

That allows you to calculate $\omega$.

Jasmine

@JohnRennie Ok

John Rennie

@Aladdin CodeClub?

Aladdin

Yes

Jasmine

I complicated it too much

John Rennie

@Jasmine :-)

Jasmine

14:52

@JohnRennie I thought finding ${\alpha}$ first then $w^2=w_0^2+2{\alpha}$ x

John Rennie

@Jasmine the problem is that question doesn't make clear what the angular acceleration is.

With a perfectly rigid string the acceleration would be infinite i.e. the mass would go from stationary to final velocity instantly.

So I don't think there's a useful way to do this by considering the acceleration.

Jasmine

@JohnRennie Okay got it :-)

John Rennie

:52065165 well in a real experiment it would, because any real string is stretchy.

user8718165

15:15

@JohnRennie sir I have a question on fermi level variation in doped sc's with temp...

John Rennie

@user8718165 yes ... ?

user8718165

@JohnRennie sir imgur.com/k6aYOve and imgur.com/ldMlaw4

@JohnRennie first is from book and second is from a source on the web....which one is correct sir?

John Rennie

It's not an area I know a lot about.

In the first image the curve for the higher dopant concentration does seem to have a maximum i.e. the Fermi level initially goes up then falls again with temperature.

And that's what the second image shows, so both could be correct. It just depends on the dopant concentration.

user8718165

@JohnRennie sir did you see by drawing a st. line? BTW its very small bump as compared to the 2nd one...

John Rennie

You need to find out why the Fermi level should initially increase as we go from absolute zero. I have to admit it's not obvious tome ...

user8718165

15:30

@JohnRennie well, thank you very much sir. Now you can take rest :-)

yuvraj singh

@JohnRennie sir hi

John Rennie

@yuvrajsingh hi

yuvraj singh

Actuality sir today I was thinking about time, what the time is, is it a material which is measurable can you describe what actually a time is @JohnRennie

John Rennie

@yuvrajsingh have a look at:

76

Q: What is time, does it flow, and if so what defines its direction?

This is an attempt to gather together the various questions about time that have been asked on this site and provide a single set of hopefully authoritative answers. Specifically we attempt to address issues such as: What do physicists mean by time? How does time flow? Why is there an arrow of ...

user8718165

16:40

Good night sir :-) @JohnRennie

John Rennie

@user8718165 Bye :-)

Transcript for

Problem Solving Strategies