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Electricity

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26 Mar 2008 12:08:09 IST

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a circular ring of uniform charge q .a small portion of it subtend angle A at centre.

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what is electric field due to remaining part of wire when small portion is removed

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NITESH BASKARAN

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Joined: 29 Sep 2007

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26 Mar 2008 12:20:35 IST

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for calculating the electric field perform the general integration but now integrate between 0 and (2pi-A)r assuming A to be in radians

Neeraj Agarwal

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26 Mar 2008 12:37:50 IST

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consider the new broken ring...

now again add a small part of charge dq(which was removed) at the broken space and a charge -dq...such that neutrality is maintained ...

now the system contains a whole ring due to which the field is zero due to cancellation...and the charge -dq due to which there is some field...

since the angle subtended is A , length of -dq is rA where r is radius of the ring...

also , the charge density of the ring is q/2.pi.r ........(charge/circumfrence)

multiplying this charge density with rA to get the value of dq...now i think u can calculate the field!

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