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Electricity

Blazing goIITian

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30 Apr 2008 22:13:37 IST

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urgent help needed , seriously!!

None

A conducting bubble of outer radius a, thickness t <<<a is charged to potential V . The bubble collapses into a droplet . The potential of such a droplet is ???

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anchit saini

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30 Apr 2008 22:15:07 IST

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Please try to give me the concept first if you don't like to type the whole thing out but please do it as quick as possible !!!

Harsh Gurung

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30 Apr 2008 22:19:39 IST

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It should be ( a/t ) V .

Sagar Vaze

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30 Apr 2008 22:20:41 IST

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q/4 pi epsilon a = V

q = 4 pi epsilon V a

E(r) = q/ 4 pi epsilon r^2 = V a / r^2

so V(r) = -qVa/r

Harsh Gurung

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30 Apr 2008 22:22:49 IST

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From potential , first find the charge , then this charge remains constant. Now the new radius is t. Then find the new potential.

saurabh

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30 Apr 2008 22:31:49 IST

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first ....

Q= CV...

now thre C = 4 pie epsilon a...

and now whn the bubble is changed into droplet..... u will get one droplet with same charge Q.. as charge wont change... remains const..

now for droplet we hve to find C' .... and put in relation Q' = C' V' = Q....now taking density of bubble and droplet to be const... as of same material....

we equate volumes....

4/3 pie[ (a+t)^3 - a^3] = 4/3 pie r'^3......... by this way we can find radius of new droplet ... using r' calculatin C'= 4 pie epsilon r'....

hehe big calculation.... but yeah... as t<< a... u can binomially expand and reduce calculation... neglecting higher powers of t....

hence put in relation of C' V' = Q...

and find new potential... V'

Cheers!!!!!!!

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