INPho-2009

1.Consider a circle of radius R.A point charge lies at a distance a from its centre and on its axis such that R=a√3.If electric field passing through the circle is ∂, find the magnitude of the charge.

16 Answers

11
sagnik sarkar ·

someone pls try!!

1
nihal raj ·

it is electric field or electric flux??

1
nihal raj ·

3 ∂. n a2 . E0 =charge where E0=epsilon zero and n=pie.....

11
sagnik sarkar ·

Sorry its electric flux.

1
sahil jain ·

actually u hav to concentrate on solid angle

1
varun.tinkle ·

a new innovative sol maybe its wrong

a point objects electric field is uniformly distribute around a sphere at the centre of whoch is the charge

all i want to say that a distance x the electic field will be uniformly distributed in a sphere of radius x at the centre of which is the charge therfore

the electirc field at a point will e constant then using this concept
E\pi a^{2}=x
thefore we know
E=x/\pi a^{2}
and total flux =E=x/\pi a^{2}*4\pi r^{2}

therfore we know E
adn therfore
q=q=12x\epsilon

where x=the sign of partial deriative

1
varun.tinkle ·

well i odnt think its innovative

shit i wont get a pinked post

1
varun.tinkle ·

anyone pls confirm it...

23
qwerty ·

23
qwerty ·

are u sure ye olympiad problem hai ?

1
bindaas ·

y not simply use solid angle !!

23
qwerty ·

bcz i don remember that !!! [3] [3]

1
bindaas ·

what u proved is solid angle concept itself :P :P

1
varun.tinkle ·

what is the right naswer n can anyone pls flaw the problem in my argument

i think it is right coz the electic field .... due to a pint charge is a spherical distribution.....

well it hink my flaw is that i hv assumed a 2d image and a circle cannot be so.....?

23
qwerty ·

surface area of a disc is not spherical ....

1
saik swargam ·

i agree with qwerty

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