ELECTROSTATICS

for 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.

sonaee see pahaleeyyy juzz complete 2000 posts....u have 1996 posts now....come on priyam u can do it........

han.. ab sona chahiye...

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[1]

priyam is feeling[13][13][13].it seems .......r u feeling [13] priyam????.....[1][1][1].........

sry i din see the above discussion

congrats Shankara... 3rd one to answer.. :P

hehe..:D

:O

v=Ed.. hota hai... maine kya likh diya...

ab sone jana hoga....dimag so gaya hai.. :P

upar edit kar dete hain.. :P

and 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..

Q.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...

yy sky???.....u dint understand the "LANGUAGE" of Q2...[12][12][12].....

i think itz C.....[1][1][1]........