1sonaee see pahaleeyyy juzz complete 2000 posts....u have 1996 posts now....come on priyam u can do it........
Q.1 Which of the following statement is true & why?
(A) Ex= Ey =Ez
(B) Ex>Ey>Ez
(C) Ex=Ez<Ey
(D) Ex<Ey<Ez
Q2. A parallel plate capacitor is charged to potential V by a source of emf . After removing the source, the separation between the plates is doubled . How will the following change electric field change on each plate potential difference capacitance of the capacitor Justify your answer.
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17 Answers
3for the first one it is c
the reason is electric field is directly proportional to crowd of the field lines or it can be said dat it is directly proportional to the no of field lines
for Q2)C=ε0A/2d
Q=CV but Q is constant
and therefore V=Q2d/ε0A and also since the electric field is uniform thru the capacitor
V=E2d
so E=Q/ε0A which says that the electric field is same as when the distance between the capacitor plates is d.
1
33han.. ab sona chahiye...
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1priyam is feeling[13][13][13].it seems .......r u feeling [13] priyam????.....[1][1][1].........
33are sorryy. kyun
its gr8 u replied....
was just joking...
ur answer is most well written.. and formatted... [1]
1hey sankara is really intelligent i think.........[1][1][1][1].........
33:O
v=Ed.. hota hai... maine kya likh diya...
ab sone jana hoga....dimag so gaya hai.. :P
upar edit kar dete hain.. :P
1and 1 is (c) acc to me
ans 2:
firstly..E=V/d and not the other way round..
now, C is prop to 1/d therefore C is halved
Q remains same after disonnecting from battery..therefore, since Q=CV and C is halved, V is doubled...
E=V/d therefore E remains constant..also E=Q/e (e=epsilon zero) and hence independent of distance..
33Q.2 E will reamain same as E is prop to Q and Q does not changes as it is isolated..
V will change..V=Ed as d is doubled.. V is doubled ...edited
capacitance will become half...
1yy sky???.....u dint understand the "LANGUAGE" of Q2...[12][12][12].....
