# Calling all Oxbridge Physicists!

Scroll to see replies haha the site worked fine for me! Have you got Real Player installed? I'll look at your problem a little later if that's ok as have to take dog for walk and then go to gym (yay 2k erg )! Meanwhile, another EM question:

If we have a spherical volume of radius a which carries a total charge Q which is distributed uniformly withint it. I get the following electric field inside the sphere:

We know that Qinside=Qr^3/a^3

thus from Gauss' Law:

Er = Qr/4*pi*Eo*a^3 where Eo (epsilon-0)

I now want to find the potential within the spherical volume, taking the zero of potential to be at infinity.

so V = - Int E.dl
so V = - Qr^2/8*pi*Eo*a^3

Is that right? I don't see how I've taken into account that V=0 at Infinity. HELP! thanks  yup, you've got it slightly wrong.....you've integrated E, but indefinitely

potential = (negative) integral of E from infinity to the point of interest
so to do that you must know E inside the sphere and E outside
Then you evaluate the integral of E from infinity -> a (using the E for outside the sphere)
then evaluate the integral of E from a -> r (using the E inside the sphere)

try it yourself and you get: Q(3a^2 - r^2)/(8pi*(Eo)*a^3) Brilliant Chloé! I especially liked "Oh, the symmetry" and "In the beginning was maxwells equations and then there was light"

Edit: have you seen the other songs by her? http://www.entersci.com/cosmic/cosmic.htm I never thought I'd hear someone sing the integal form of stokes theorem! i know this isnt technically physics but what the hey:

10 red balls in a bag and 5 green ones. Question says "What is the most probable score after 12 draws (replacing the ball each time) and what is the expectation value of this score?"

see i would have thought the most probable score is the expectation, but then I reckon it's asking what is the most likely combination of reds and greens...i.e. which nCr is biggest. But then what does it mean by expectation value? It could be because the expected value won't actually be a possible score like when you roll a dice, the expected value is 3.5 but the probability of rolling a 3.5 is obviously zero yeah but it's saying "expectation value of this score" Chloé, that was the best video ever in the whole world that I have ever seen. Ever. If I had any rep points to give, I would give them all to you now. Pure, pure genius. I would also give all my rep to whoever made the video.

Although where the hell does this 1/c come from? Was she working in Gaussian units or something?

OK...it's possibly a little geeky, but incredible all the same. I'm off to spread the joy by posting that everywhere I can. that is so terrible it's good. What have I become!? The women who made it calls herself the Physics Chanteuse! Willa
yup, you've got it slightly wrong.....you've integrated E, but indefinitely

potential = (negative) integral of E from infinity to the point of interest
so to do that you must know E inside the sphere and E outside
Then you evaluate the integral of E from infinity -> a (using the E for outside the sphere)
then evaluate the integral of E from a -> r (using the E inside the sphere)

try it yourself and you get: Q(3a^2 - r^2)/(8pi*(Eo)*a^3)

Thanks Will, I agree with your answer; so even though I only want the potential within the volume, I need to consider the eletric field outside?

I PROMISE to look at your question in a bit...although I must admit I'm possibly even as bad at Vector Calculus as EM...so I wouldn't hold out any hope ! Nick/Phil/Andy please can you look at his vector calculus question, which I think is now at the bottom of the last page.

Also, glad everyone appreciated the video ...Andy, perhaps it's time to reveal the 'Matrix' photo? I don't think Nick or Phil have seen it (afraid it's an Oxford joke Will that you won't get)! I get the same answer as you for that line integral, Will - have faith in yourself! I don't know why you didn't feel comfortable with the answer - it is dimensionally correct...

Ahh yes, Chloé....what is the matrix, exactly? I still don't know. I only realised a couple of days ago that whoever did it spent the time giving him a suitable waistcoat too! I don't think Neo wore that. V.quick question guys (I know I always say that!!!).

We have a dipole and I know how to prove that:

Potential = (z*q*d)/(4*pi*Eo*r^3)

I also know that the Diploe Moment of 2 charges, is:

p=qd

however, I don't see why that gives us

Potential = (p*cos[theta])/(4*pi*Eo*r^2)

Clearly we're saying that qd = p*cos[theta]*r but I don't understand why that is. I'm guessing the cos is because we have a dot product but the r?! Thanks...and sorry to be annoying and ask so many questions (but at least it's good revision for you guys!) p points from the -q to the +q: a distance of d.
r points from the centre of the dipole (midpoint of p, d/2) to the point A.
[t] is the angle between them

V = q / (4pi{e0}r)

So the potential at A is the sum of the potential from both the charges, using the cosine rule to find the distance, r, to each charge. Notice the (-d/2) in the positve charge's term because it's cos(pi - t) = -cos(t).

V = (q / (4pi{e0})) * ({(r^2+(d/2)^2-d*r*cos[t]}^(-1/2) - {(r^2+(d/2)^2+d*r*cos[t]}^(-1/2))

Take out r^2 from each of the terms....

V = (q / (4pi{e0}*r)) * ({(1+(d/2r)^2-(d/r)*cos[t]}^(-1/2) - {(1+(d/2r)^2+(d/r)*cos[t]}^(-1/2))

You've got (1 + x)^(-1/2) in the terms, so use binomial theorem and it drops out. Worzo
I get the same answer as you for that line integral, Will - have faith in yourself! I don't know why you didn't feel comfortable with the answer - it is dimensionally correct...

Ahh yes, Chloé....what is the matrix, exactly? I still don't know. I only realised a couple of days ago that whoever did it spent the time giving him a suitable waistcoat too! I don't think Neo wore that.

well then i have a problem with the next bit. It says:

well Grad(p) I got as (0,z^2,2yz) but I dont see how it helps!

and enough of your oxford in jokes...i feel left out enough as it is Willa
can some please evaluate the line integral around the circle: x^2+y^2=r^2 (z=(z0) i.e. a constant) of the following field:

U = (yz^2, yx^2, xyz)

I get -pi*r^2*(z0)^2 but that just feels wrong, so can someone have a go please

yer, that's right shiny
yer, that's right

tehn can you have a stab at the next part (2 post above). cheers You're answer to that end bit is definitely correct. Perhaps there is no connection! lol Also, what would you guys write as your answer if you saw the following question:

Q. Why are impedances usefully represented by a complex number?

Would you just say it's because the complex number carries information regarding both the magnitude and phase, which is important for understanding an AC circuit? Hoofbeat
You're answer to that end bit is definitely correct. Perhaps there is no connection! lol

well there has to be a connection...somewhere... and I reckon your answer to complex number impedence is correct. I'd possibly phrase it: "Because circuits containing resistors capacitors and inductors running in a steady state have voltages and currents varying sinusodially with the same frequency but different magnitudes and phases. Complex notation contains all of this information". But i'm sure your answer is adequate Thanks Will. Have you tried asking your vector calculus question on Physicsforums? Although, I'm not a fan of that site, as I find the answers sometimes misleading, there are a lot of knowledgeable people on there, and I'm sure someone can help!

Btw, Andy you know the question you responded to earlier regarding the dipole, I've just seen a v.simple way to get from:

Potential = (z*q*d)/(4*pi*Eo*r^3)
to
Potential = (p*cos[theta])/(4*pi*Eo*r^2)

It's because on my diagram, z = rcos[theta], which substitutes in perfectly to give the second answer I'm sure you way is probably more elegant, but I like Feynman's derivation of the potential of a dipole in cartesian coordinates, and then it's simple to convert to spherical etc.

Finally, Shiny, you've been hiding (well I imagine working too hard ) - I haven't seen you on msn etc all day!