1plsss help...............
1.total charges on all capacitors=q1.total charge moved through the battery=q2.q1 and q2 add up to 16μC. and differs by 2μC.then x=?
a.1/3 b.1/2 c.2/3 d.3/2
2..an infinite sheet has asurface charge density of 2\times 10^{-8}C/m^2.the work done by the external energy (Joule)in taking a unit positive charge from distance 100cm to 10cm towards the sheet approx. is a.10 b.100 c.1000 d.2000.
3.the flux through the curved surface of a cone due to a charge placed at the base centre is @.when the same charge is placed at one corner of a cube ,the electric fluxes through 2 adjacent surfaces of the cbe are @1& @2.THEN @1+@2=?
a.@/4 b.@/8 c.@/12 d.@/16.
4.A parralel plate capacitor of 1.11μF is designed using a material of dielectric constant 22.2 and breakdown strength 3.33\times10^6V/m.to make this capacitor withstand ap.d of 444V accross the plater,the value of the distance beween the plate should be atleast
a.5.8\times10^-7m b.6\times10^-6m c.6.66\times10^-5m d.7.2\times10^-5.
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5 Answers
1i may make a calculation mistake ... coz i think the ans shud be @/12...
in haste... will come later..
3see the first part its placed on the bottom of a cone ... so only half the flux will go thru the curved surface .......................therefaore ..... we have
q/2E = @ [ takin epsilon as E :P]
there are 24-8 phases thru wich charge can flow .............
q/16E flux thru each side ,,,,,
therefore @/8E ... shud be @1 + @2 ....
thus @/4 = answer ...!!!!!!!!!!!!
NOTE : the eight phases in line with the charge no flux pases because angle betw area vector and flux =pi/2 rad
3for second question by drawing a cylindrical gaussian surface thru the sheet we see that that E field = sigma / 2E [E = epsilon]
since E = constant ... ∫Edr = Er ....
u can take it frm here ....;)