Calling all Oxbridge Physicists!

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)

Reply 142

Bezza

2

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!

Reply 143

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?

Reply 144

Bezza

2

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

Reply 145

yeah but it's saying "expectation value of this score"

Reply 146

Morbo

17

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.

Reply 147

that is so terrible it's good. What have I become!?

Reply 148

Bezza

2

The women who made it calls herself the Physics Chanteuse!

Reply 149

OP

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 :frown:

! 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 :p:

...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)!

Reply 150

Morbo

17

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.

Reply 151

OP

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!)

Reply 152

Morbo

17

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.

Reply 153

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:

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!

and enough of your oxford in jokes...i feel left out enough as it is

Reply 154

shiny

12

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

Reply 155

shiny

yer, that's right

tehn can you have a stab at the next part (2 post above). cheers

Reply 156

OP

You're answer to that end bit is definitely correct. Perhaps there is no connection! lol

Reply 157

OP

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?

Reply 158

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... :frown:

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

Reply 159

OP