713. It is a good question. Conserve flux through a cross section. Using
q1 (1-cos a)2ε0 = q2 (1-cos b)2ε0
Where, a= x1/2 , b = x2/2
1) A spectroscopic instrument can differntiate 2 close wavelengths λ and λ+Δλ if λΔλ = 8000. The instrument is used to study spectral lines of balmer series of hydrogen. find approx how many lines are distinguished by the intrument.
2) A charged particle of charge q is revolving in a circle of radius R of time period T = 9 sec. A point P at distance 2R from centre of circle in same plane records magnetic field . find min. time interval betqween 2 succesive times at which mag. field is 0 at tht point.
3) An electric field line emerges from the point charge q1 = 36 C at an angle x1 to the straight line joining both charges. if same lined enters the negative charge at x2 . given x1 = 60 degree and x2 = 74 degree
then find magnitude of the other charge
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13 Answers
7171I think question 2 has been asked earlier by aditya.. Try looking for them
71Okay. So what i mentioned isn't clear enough ?
See with figure :
Now, we know that flux due to a charge in that " conical " envelope is as mentioned earlier in my post. [ You can also derive it using the fact that total flux = q/ε0 , then using Solid angle concept obtain it for some half angle a]
1hmm..i've never seen tht formula earlier..
also i dun understand "conserve flux through cross-section" for that matter let's take an infinite sheet instead of a small cross section so by "conserving flux" we get Q1/2ε0 = Q2/2ε0 which gived us Q1 = Q2 :O
71Since you are asked the relation between x1 and x2. It is given that a line which emerges from A goes to B.So you can safely say that the flux(ie., other lines) beneath it, certainly reach to the other charge. So you would conserve that only between that.
21In the system of two charges given,flux goes from charge 1 to 2.However some will definitely escape[draw the diagram of flux lines].The amt. of flux loss will depend on the charges.But if you consider at angles less that x1 flux has to be conserved.Therefore we can apply flux conservation here at angles less that x1,as it is mentioned that flux at this angle reaches the other charge.
For the solid-angle concept.Consider a cone with semi-vertical angle α.Now it says that area(part of sphere with centre at vertex of cone)=2πR2(1-cosα)
To prove area=2π(1-cosα),we integrate from 0 to α:∫2πRsinθ*R∂θ=2πR2(1-cosα)
Therefor solid angle=2πR2(1-cosα)/4πR2=(1-cosα)/2
Now,utilize flux conservation.