Recent content by ku1005

Electron...achievable resolution? Homework Statement An electron is confined inside a sphericial region of diametre 1.5nm.The speed of the electron is measured. Which of the following is not an achieveable resolution for the measurement? A +/- 9.7*10^3 ms^-1 B +/- 3.86*10^4ms^-1 C...

thanks for all your help! greatly appreciated...deserve a medal for the length of this convo lol

oh and just quickly...of course your correct... when using only the 2 faces, of which, both have surcae area of 3*10^-6 m^2, then the calculation gives exactly the answer... ie (1.08*10^-16) / ((8.85*10^-12)*(2(3*10^-6)^2)) = 0.68V/m!

Doc Al your a champ!...i really appreciate all your help! You highlighted the reason why i have gone wrong from the start, i will show you my thinking: I imagined the set of charges within the slab as a point charge at the centre of the Guassian Surface in this manner...

um my cube was (3mm)^3 sorry

Okay...im getting closer with your help...which is great!:) This time i followed your suggestion, and cough cough attempted again. I understand the equivalent 1mm from both the top and bottom of the slab now. Heres the question, do also make it 1mm from the other sides? Or better...

i think I understand why my approach won't work, I don't think I am able to choose a "symmetical" guassian surface which encloses all the charge of the slab, and which also has "point P" on its surface. What I previously selected was a smaller cube within the slab itself, but in doing so, I...

im sorry , i don't quite understand how you consider it is composed of planes of charge...My understanding (and I am in no way saying I am correct,just trying to show you my thinking) is that the electric field radiates outwards in every direction from a charge. HENCE...what I was meaning, is...

Guassian Surface Decision? Homework Statement http://img412.imageshack.us/img412/4056/electroquestionhd7.png Homework Equations http://img337.imageshack.us/img337/6898/guasslawdd2.png The Attempt at a Solution My thoughts/attack for the question is first to find the charge...

im an idiot... seen as Potential Energy Difference = Work Done And we know Potential = Energy per unit charge, it makes sense that the potential difference of 1050V multiplied by the charge concerned (ie electron) will give the Change in Potential Energy and thus the work done. Is this...

Homework Statement Two conducting spherical shells are concentric, with radii of 0.70 m and 1.60 m. The electric potential of the inner shell, with respect to the outer shell, is +1050 V. An electron is transported by an external force from the inner shell to the outer shell. The work done by...

just thinking now, it because the amplitude of light is unaffected by the intensity of the light, and therefore, throughout the 3 polaroids, the amplitude of the light will simply change via: E-> first Ecos(alpha) through second Ecos(alpha)cos(beta-alpha) through third We only need...

Homework Statement Unpolarised light is shone onto a set of three transmission Polaroid sheets. The intensity of light after the first Polaroid sheet is measured to be I = E2. The transmission axis of the second and third Polaroid sheets are at angles of alpha and beta with respect to the...

yes i understand now, thanks for your help!

mmm just thinking...not sure why it would be (Distance / Time )* 2pi, becasue that makes no sense. It should be (Distance/Wavelength)*2pi which would therefore give a percentage of the complete cycle and hence an angle!- opr phase diff...is this thinking correct??