Calling all Oxbridge Physicists!

OP

Willa

but what happened to the q*d then?

we substituted the p=qd

Have fun at the pub Will, I'm sure you deserve it as you've been working hard. Don't worry about the rowing next week - I don't know if you have collections when you go back (or what you call them in tabland :rolleyes:

) but you've been able to answer pretty much everything on here, so I'm sure you'll do fine in your tests when you go back and you'll definitely be fine in your prelims! Besides, you should relax and do something you enjoy (ie. rowing, although I'm sure it'll be tiring, rather than relaxing! hehe) before you get back into working hard for a whole term - a happy mind means you'll work better!

Anyways, one more question guys (last one for tonight I promise, as once I've done this I'm stopping work!!!)...been doing a past exam paper and I'm not sure how to do this question, as never met one like this before (we don't use the lecturer's problem sheets, only our tutors!):

Right, we have a bog standard dipole with a potential of pcos[theta]/(4*pi*Eo*r^2). Now we stick the dipole in a non-uniform electric field along z-axis (dipole also along z axis) of magnitude E=kz. What force does the dipole experience? I know that F=Grad[p.E] but how do I continue? Thanks

Reply 162

Morbo

17

Ah ha. You may be having problems because this is where Kraus was silly and was wrong in his notes. Actually, F = (p.del)E. If the force was given by something to do with p.E, then this would imply there could never be a force on a dipole lying perpendicular to the E-field lines, which is not true!

Reply 163

well you forget that I dont do just physics!

and as a quick stab at your question:

when it says "force on the dipole" are you talking about the vector sum of the force on the two charges. And Force = qE, so find E at the two points. For example for a charge +q, at +z, E=kz + E due to other point charge...which you can find from the standard equation.

i dont think that question asks for any fancy dipole jiggery pokery...merely to use the fact that the dipole has a standard potential to deduce the positions of the two point charges.
Although then again I may be completely wrong....enough physics for one day.

Reply 164

OP

Worzo

Ah ha. You may be having problems because this is where Kraus was silly and was wrong in his notes. Actually, F = (p.del)E. If the force was given by something to do with p.E, then this would imply there could never be a force on a dipole lying perpendicular to the E-field lines, which is not true!

ARGHHH Annoying Kraus!

Thanks for the suggestion Will, but I don't think that's how we're expected to do it. Also, you should be at the pub!!!

Andy (I know you've disappeared for the moment, but whenever you're back), how do I progress using the methods we were supposed to be taught? I don't know what the components of p are though (probably me being ignorant!)?

Reply 165

shiny

12

Willa

well then i have a problem with the next bit. It says:
Calculate Grad(p) where p=yz^2 and hence explain your answer
well Grad(p) I got as (0,z^2,2yz) but I dont see how it helps!

I can't see the connection? :confused:

Er ... ???

Reply 166

Morbo

17

I might be misinterpreting how you've phrased the question, but it just seems like a simple plugging in the values question.

If E = kz, then there is only E-field lines in the z-direction. You say the dipole is also lying along the z-axis, so p is entirely in the z-direction.

F = (p.del)E = (p * d/dz)(kz)

|F| = pk in the z-direction

Reply 167

Hoofbeat

Thanks for the suggestion Will, but I don't think that's how we're expected to do it. Also, you should be at the pub!!!

by the time you made this post I was daaan pub. THE ACE OF SPADES!.....hmmm i'm drunk. I shall no doubt delete this in the morning

Reply 168

OP

ace of spades? ummm

Ok Andy, thanks for the brief explanation - tht was exactly what I needed!

Also, one of the other Lincoln Physicists found this last night, and I thought you might enjoy it (it's not as good as the videos in my opinion):

The Physicists' Bill of Rights
We hold these postulates to be intuitively obvious, that all physicists are born equal, to a first approximation, and are endowed by their creator with certain discrete privileges, among them a mean rest life, n degrees of freedom, and the following rights which are invariant under all linear transformations:

To approximate all problems to ideal cases.
To use order of magnitude calculations whenever deemed necessary (i.e. whenever one can get away with it).
To use the rigorous method of "squinting" for solving problems more complex than the addition of positive real integers.
To dismiss all functions which diverge as "nasty" and "unphysical."
To invoke the uncertainty principle when confronted by confused mathematicians, chemists, engineers, psychologists, and dramatists
When pressed by non-physicists for an explanation of (4) to mumble in a sneering tone of voice something about physically naive mathematicians.
To equate two sides of an equation which are dimensionally inconsistent, with a suitable comment to the effect of, "Well, we are interested in the order of magnitude anyway."
To the extensive use of "bastard notations" where conventional mathematics will not work.
To invent fictitious forces to delude the general public.
To justify shaky reasoning on the basis that it gives the right answer.
To cleverly choose convenient initial conditions, using the principle of general triviality.
To use plausible arguments in place of proofs, and thenceforth refer to these arguments as proofs.
To take on faith any principle which seems right but cannot be proved.

Reply 169

Bezza

2

I wanna see the matrix photo! Is it of hans?? The oriel physicists are obsessed with him and even made a yard stick after his data analysis lectures!

Reply 170

OP

lol, no it's of Dieter!!! Unfortunately dad's taken my laptop this morning so we can get the network key setup for the ADSL, so I'm on my desktop and I don't have the photo on here. I'll see if I can find it on the web, if not I'll send it to you over msn this evening!

What's a yard stick? :confused:

(shows her ignorance)

Reply 171

OP

Hey, guys another question! Am doing a method of images question, and have got the E-field to be (it's a point charge directly above a conducting plate):

E = (-q2d)/(4*pi*Eo*d^3)

which can obviously be cancelled to give

E = -q/(2*pi*Eo*d^2)

Now, I want to find the force is experiences. Now F=qE therefore, I would have just anticipated that F = -q^2/(2*pi*Eo*d^2)

However, in Feynman, the answer is F = q^2/(4*pi*Eo*[2d]^2). I can't see where he got the [2d]^2 from, nor how he's still got the 4pi on the bottom! He also didn't get a minus sign, but I've checked my signs and I'm certain that it's correct and it makes sense that the positive charge will be getting pulled down towards the plate (or at least it does in my mind!). Thanks

Reply 172

as far as i am aware the yardstick is just a fancy word for a ruler....and is most often 1 yard long?

and sorry, last night me and my friends developed an obsession with motorhead's "the ace of spades" for no particular reason. I think it was cos we had had enough of doing hard core gangster style dancing to RnB (by hard core gangster style dancing that means us taking the piss out of the way they dance on their videos these days)...well at least complete strangers thought we were funny, but they were probably laughing at us rather than with us!
But yes we demanded the ace of spades be played and they didnt, so we were very upset.

anyways, can one of you kindly folks take a gander at the following please:

Evaluate the integral:

Unparseable latex formula:

\bigint_0^\infty \cos (\alpha x) \exp^{- \beta x} dx

By substituting for:

\cos (\alpha x) = Re(\exp^{j \alpha x})

Basically I have tried this and I was wondering if you can put that into the equation and then take the "Re" part outside the integral....Either ways, can someone have a go at evaluating this. The answer I get his:

\frac{-\beta}{\alpha ^2 + \beta ^2}

The next part then says: Hence evaluate:

Unparseable latex formula:

\bigint_0^\infty x\cos (\alpha x) \exp^{- \beta x} dx

To do this I wrote that:

Unparseable latex formula:

\bigint_0^\infty x\cos (\alpha x) \exp^{- \beta x} dx = -\frac{d}{d\beta}\bigint_0^\infty \cos (\alpha x) \exp^{- \beta x} dx

and hence differentiated my answer to the last part, to obtain:

\frac{\alpha^2 - \beta^2}{(\alpha^2+\beta^2)^2}

is this the correct answer?

and wohoo i'm getting the hang of this latex stuff!

Reply 173

Hoofbeat

Hey, guys another question! Am doing a method of images question, and have got the E-field to be (it's a point charge directly above a conducting plate):

E = (-q2d)/(4*pi*Eo*d^3)

which can obviously be cancelled to give

E = -q/(2*pi*Eo*d^2)

Now, I want to find the force is experiences. Now F=qE therefore, I would have just anticipated that F = -q^2/(2*pi*Eo*d^2)

However, in Feynman, the answer is F = q^2/(4*pi*Eo*[2d]^2). I can't see where he got the [2d]^2 from, nor how he's still got the 4pi on the bottom! He also didn't get a minus sign, but I've checked my signs and I'm certain that it's correct and it makes sense that the positive charge will be getting pulled down towards the plate (or at least it does in my mind!). Thanks

are you trying to find the E at the point where the charge is placed? Because if so, what I think feynman has done is said that the point charge experiences a force that is equivilent to an image point charge pulling at it

The image point charge will be a distance of 2a from the real point charge, and has charge -q. So using simple coloumbs law you get feymans answer

Reply 174

shiny

12

Willa

Basically I have tried this and I was wondering if you can put that into the equation and then take the "Re" part outside the integral....Either ways, can someone have a go at evaluating this. The answer I get his:

Yer cos integration is a linear operation.

\int e^{j \alpha x} e^{\beta x} dx = \int ( \cos (\alpha x) + j \sin (\alpha x) ) e^{\beta x} dx

= \int \cos (\alpha x) e^{\beta x} dx + j \int \sin (\alpha x) e^{\beta x} dx

So as long as the cos and sin integrals are real then you can take the Real part of the integral.

Yes, I think the differentiation step is fine, although I haven't checked the algebra but the method looks right.

Reply 175

wooo.....i cant believe it i seem to have got another question right!?

Reply 176

sorry, but i have another annoying surface integral to do, i think i know how to do it, but as usual i need some re-assurance:

I have to evaluate the surface integral F.dS over the surface: r=(acos[phi]sin[theta],bsin[phi]sin[theta], ccos[theta]) (over all angles)
where F=r=(x,y,z)

so do I just take F to be: (acos[phi]sin[theta],bsin[phi]sin[theta], ccos[theta]) and then convert dS into a form which contains d[theta]d[phi] and evaluate over all angles?

and to find dS the question says to use:

dS = \frac{\partial r}{\partial \theta} \times \frac{\partial r}{\partial \phi}d\theta d\phi

(I presume that's a cross product in there)
for that I got:

dS=(cb\sin^2\theta\cos\phi, ac\sin^2\theta\sin\phi, ba\cos\theta\sin\phi)d\theta d\phi

can someone confirm that as correct?
so then do I just dot product that with F=(acos[phi]sin[theta],bsin[phi]sin[theta], ccos[theta])? and integrate over all angles

Reply 177

OP

Willa

are you trying to find the E at the point where the charge is placed? Because if so, what I think feynman has done is said that the point charge experiences a force that is equivilent to an image point charge pulling at it

The image point charge will be a distance of 2a from the real point charge, and has charge -q. So using simple coloumbs law you get feymans answer

I'm really not sure what I'm trying to do! Here's the exam question:

"Using the method of images, find the force between a charge q and a large flat conducting plate at zero potential when the charge is at a small perpendicular distance d from the plate."

So I worked out the potential and then the electric field (-dV/dz) and got the value that I quoted before.

I get what you're saying about just saying it's coulomb's law and you can get Feynman's answer simply that way, but surely my method (up to finding E-Field) will work and give the correct answer. Any ideas why I can't get the correct answer when I use the electric field?

Reply 178