Calling all Oxbridge Physicists!

but hold on, if you have a point charge next to a conducting plate, the electric field is identical to a situation where you have a point charge near a point charge of opposite charge. So surely you can just use field strength and coloumbs law as if there really was a charge there! Just like you have used potential...surely if the potentials are equivilent then so are the E and F values.

Reply 183

shiny

12

Willa

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

looks alright to me

Reply 184

thanks alot...glad someone isnt afraid to tackle these horrific looking questions i'm posting up these days

neways, best be off for my last erg before training camp

Reply 185

OP

WOAHHHHHHHHHHHHHHHHHHHHHHHHHH my post is causing havoc! Right:

1. Thanks Will for the derivation, although, as Andy said it looks over complicated (esp for someone as 'dense' as me :p:

!!!).

2. Thanks also to Andy for trying to help!

RIGHT! I think the only way to help me is to look at our lecture notes, so Will you can take a break :p:

Andy, if we look at the notes good old Kraus-y gave us, then under image charges he derives that E= - qd/(2*pi*Eo*(r^2 + d^2)^3/2 which I completely agree with! So if we're considering the e-field directly below the charge, thus r=0 and E = -qd/(2*pi*Eo*d^3). How do we then get the force? Surely it's just F=qE, which implies F = - q^2/(2*pi*Eo*d^2). HOW COME EVEN THOUGH WE'VE USED THE DERIVATION WE DON'T GET THE SAME FORCE WE'D GET IF WE USED COULOMBS LAW? ie, why do we have only d^2 instead of (2d)^2 and how come we have (2*pi) on the bottom instead of (4*pi)?

Cheers guys, I know I'm being annoying but I don't see what's wrong with what I'm doing

Reply 186

it looks like the E you are calculating is the E at the surface of the plate conductor. But the force on the point charge is = q*E at the position of the point charge. And the point charge is NOT on the surface of the plate....it is slightly away from the plate. You must find the E at a distance "d" from the plate...the E you have is the E at the point on the plate directly below the point charge. hence why you dont get the right answer - you're using the wrong E

now i'll shut up and go erg

Reply 187

OP

ok thanks Will, that was all I wanted to know!!! :biggrin:

Hope the erg wasn't too painful...I'm feeling the affects after my attempt at a new 2k PB (unsuccessful I might add, but I always seem to erg slower on the one at the gym than at college anyways!) and my legs now hurt (although that could possibly be from the squats I did afterwards!). Anyways, I've procrastinated FAR too much today...so work calls.

Reply 188

woooooo 7:10.4 baby! New PB by a whole 10 seconds. And what's amazing is that I did that not at max pressure! I was orginally planning to do 3x2000m with 5 mins between each, but go at each at 90% pressure (so start them with a split of 1:49-1:50 and then work from there). But I started the erg and every time I try and go slow when doing a piece, I just cant manage it. So anyways I set off and found myself sitting rather happily at about 1:46. After 1000m I found I was averaging about 1:47 and thought "I'm pretty tired now, but bloody hell I'm well on target to smash my PB....screw this I'm going for it" and I put in everything from then on and got 7:10.4!! So that means not only did I get that time without putting in max effort all the way through (i didnt feel sick afterwards which is a good indicator i didnt try my hardest) but I normally do better when I have encouragement from my teammates around me. Oh how I would love to get near 7:05!!! Where's your PB at atm then?

ok yes, back to revision!

Reply 189

OP

lol, well done! My PB is currently 8:26, which was only at a rate of 26 and I was quicker than some of the current people in the first boat who were rating 30 (I was doing the trials at the same time)! Yesterday I tried a new technique of trying to keep the same pace all the way through, and I got 8:34 (which is still good compared to others in my boat, so I'm hoping my seat's safe at the trials), but when I go next week I'm going to use my old technique of going off at a ridiculous pace (head off at my 500m pace of 1:52-1:58) as I find that works for me! I know people say you should pace yourself and then sprint at the end, but by the end I'm too tired, so I'd rather go hard off the start (plus in bumps racing that's what usually happens as everyone tries to bump). Never mind; like I said I always find I do erg much slower at home (I swear it's the old erg!!). I did a 2k just before I went back after xmas and got 8:56 at the gym and then within 5days I got 8:26 in the 2k trials (30seconds off a PB within 5days without any other exercise seems a bit unbelieveable!!!!).

Reply 190

OP

Right, back to work and yet another problem (I blame our stupid lecturer's notes). As it's a problem I'm having directly with the lecture notes, I'm not going to write it all up (as I'd have to draw the diagram), so perhaps Phil/Andy/Nick could explain why he does what he does (unless Will fancies showing the derivation they've learnt?)

B-Field of a Solenoid Using Biot Savart Law [reference: 2.2.3]

Ok, I get how Kraus got Bz = Mu-o*I*Integral [Na^2]/[2*l*(a^2 + z^2)^3/2] dz

But how the hell did he use that to say that Bz = Mu-o*I*(1/2)*(cos[alpha] - cos [beta])*N/l

I get what alpha and beta are, but how did that expression come from the above one?!?!?!?!?!

Similarly:

B-Field of Straight Wire [ref: 2.2.5]

We've got the magnetic field in phi direction:
dB = (Mu-o*I*dl*sin[theta])/(4*pi*r^2)

he then says sin[theta]=a/r and tan[theta]=a/L which it obvious, but then how does he use them along with dl->d[theta] to get B=Mu-o*I/2*pi*a

GRRRR I hate his notes so much!!!!!!!!!!!!!!!!!!!

Reply 191

shiny

12

I get this feeling that even the crappy notes I had in 1st year are better than yours. Pity I burnt them after 1st year exams :biggrin:

Reply 192

well if you got 8:34 on your crappy old erg, then surely you're gonna get 8:04 on the college ergs if you pace yourself! You shouldnt underestimate pacing...what I do is I decide what I want my average split to be over the whole thing, then my race plan is: first 500m go at about 2-3 seconds under that split....even less if you feel comfortable, but you shouldnt feel you're working hard to maintain it. Then second 500m drop your split to just below your target split...just half a second or 1 second will do. Then in the next 700m you can let it slip to 2 seconds slower than your target split...and then in the last 300m just go crazy. So far it's helped me reach my target times pretty well (and you tend to beat them as well cos in the first 500m you tend to do a little better than just 2-3seconds faster...but that's what you should aim for)

but as you said, whatever works for you!

Reply 193

OP

Cheers thanks Will! I'd be quite happy with 8:20 at the moment! I know I should try and learn to pace myself but my brain doesn't like it and after the first 1k I'm like "NOOOO I WANT TO GIVE UP" and so if I haven't gone quick to begin with I just never pick up and get quicker! Originally I'd planned the first 500m quick, then down for 500m, then up for 250, down for 500m and up like mad for the last 250 (I hate having 300m left to go, that's the moment when I feel most like giving up, even when I'm doing 500m, it's the psycological moment for me!). I also have a phobia of keeping the rating high (32) on the erg as I don't believe I'll be able to hold it, so I drop the rate quite quickly and then never get it up again! I know I'm capable of it as we rated 32-34 for the 2days when we had to row-over in bumps (2k) but then because it was competition I just kept going and driving with my legs (I was sick after our first row over as soon as we crossed the finish line! oops!) On the last day, I was grinning with the last 500m to go as we rowed past the boathouses as I could hear everyone cheering and although the rest of my crew said they were in agony, I knew it was the last 500m of Torpids and for me it was my boat race and I was just going for it (the others told me they ignored our coxes call to "power-20" :eek:

as we weren't going to catch the crew ahead of us and the other crew were too far behind us).

Grrr...back to more ****** EM

Reply 194

Hoofbeat

Right, back to work and yet another problem (I blame our stupid lecturer's notes). As it's a problem I'm having directly with the lecture notes, I'm not going to write it all up (as I'd have to draw the diagram), so perhaps Phil/Andy/Nick could explain why he does what he does (unless Will fancies showing the derivation they've learnt?)

B-Field of a Solenoid Using Biot Savart Law [reference: 2.2.3]

Ok, I get how Kraus got Bz = Mu-o*I*Integral [Na^2]/[2*l*(a^2 + z^2)^3/2] dz

But how the hell did he use that to say that Bz = Mu-o*I*(1/2)*(cos[alpha] - cos [beta])*N/l

I get what alpha and beta are, but how did that expression come from the above one?!?!?!?!?!

Similarly:

B-Field of Straight Wire [ref: 2.2.5]

We've got the magnetic field in phi direction:
dB = (Mu-o*I*dl*sin[theta])/(4*pi*r^2)

he then says sin[theta]=a/r and tan[theta]=a/L which it obvious, but then how does he use them along with dl->d[theta] to get B=Mu-o*I/2*pi*a

GRRRR I hate his notes so much!!!!!!!!!!!!!!!!!!!

I cant help you too much cos as you said i dont know what he's using for these signs, but if it helps, here's B Fields....The Cambridge Way:

Take a long solenoid and place a rectangular amperian loop of length l around one side of the solenoid. If the solenoid has N turns and is length L then this means the rectangular loop will contain: INl/L current through it. By Ampere's law: integral B Field around loop = Mu-0*INl/L
If the solenoid is long then the only contribution to the B-Field is along the inside of the solenoid (cos the outside of the solenoid has 0 magnetic field cos you can take that side of the recantagle to infinity and it's still valid...hence is 0). So Bl = Mu-0*INl/L
or B = Mu-0*IN/L

Bio-savart method to find field around straight wire:

dB = (Mu-o*I*dl*sin[theta])/(4*pi*r^2)

you need to integrate that from -infinity to +infinity over dl
but r = a/sin[theta] and L = a/tan[theta]
so dL/d[theta] = -asec^2[theta]/tan^2[theta] = -a/sin^2[theta]

so plug your new r and dL/d[theta] in to dB (and using the fact that dL = dL/d[theta] * d[theta])

dB = (Mu-o*I*-a*sin^3[theta])/(sin^2[theta]4*pi*a^2) d[theta]
= -Mu-o*I*sin[theta]/(4*pi*a) d[theta]
integrate that between the limits which are pi->0 and you obtain the desired answer!

Reply 195

KuinKra

1

Hoofbeat

Right, back to work and yet another problem (I blame our stupid lecturer's notes). As it's a problem I'm having directly with the lecture notes, I'm not going to write it all up (as I'd have to draw the diagram), so perhaps Phil/Andy/Nick could explain why he does what he does (unless Will fancies showing the derivation they've learnt?)

B-Field of a Solenoid Using Biot Savart Law [reference: 2.2.3]

Ok, I get how Kraus got Bz = Mu-o*I*Integral [Na^2]/[2*l*(a^2 + z^2)^3/2] dz

But how the hell did he use that to say that Bz = Mu-o*I*(1/2)*(cos[alpha] - cos [beta])*N/l

I get what alpha and beta are, but how did that expression come from the above one?!?!?!?!?!

Similarly:

B-Field of Straight Wire [ref: 2.2.5]

We've got the magnetic field in phi direction:
dB = (Mu-o*I*dl*sin[theta])/(4*pi*r^2)

he then says sin[theta]=a/r and tan[theta]=a/L which it obvious, but then how does he use them along with dl->d[theta] to get B=Mu-o*I/2*pi*a

GRRRR I hate his notes so much!!!!!!!!!!!!!!!!!!!

Well, the second has been answered by our cambridge friend, so I'll fix the first one for ya.

With a long solenoid, it's made up of many cocentric rings. You want to sum up all the contributions over the length of the solenoid.

For one loop:

B = (mu-o*I*a^2)/2*(a^2 + z^2)^1.5, where z is the distance over the solenoid.

Therefore, dB = [(mu-o*I*a^2)/2*(a^2 + z^2)^1.5] dN, where N is the increment in hoop number.

dN = ndz

Therefore, dB = [(mu-o*I*n*a^2)/2*(a^2 + z^2)^1.5]dz

atan(theta) = z, where theta is the angle between the normal of the hoop and the distance r to the point away from it.

Therefore asec^2(theta) d(theta)= dz

And remember sec^2(theta) = (a^2 + z^2)/a^2

Substituting all in:

dB = [(a*mu-o*I*n)/2(a^2 + z^2)0.5] d(theta)

And a/(a^2 + z^2)^0.5 = cos(theta)

So the final term is:

dB = 0.5mu-o*I*ncos(theta) d(theta)

And you integrate over the two angles that describe the solenoid (alpha and beta, or whatever).

To get the expression Kraus used, merely substitute phi = pi - theta. I've resolved from a different angle by mistake and I can't be bothered to type it all out again. But it still works niftily.

Reply 196

OP

Thanks both to Will & Phil :biggrin:

. Why did Kraus have to leave out so many stages...not all of us are capable of 'guessing' what he did!!! <grumbles>

Reply 197

stuck trying to do a bit of fourier series....i've got it down to trying to evaluate:

Unparseable latex formula:

\displaystyle \bigsum \sin(nx)\frac{1}{\pi}\int_{-\pi}^\pi\sin(ax)sin(nx)dx

where a is not an integer
any ideas how to proceed?

Reply 198

OP

afraid we don't cover fourier series til 2nd year, although I imagine Phil's been teaching himself some!

Reply 199