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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jul 2007 12:22:45 IST
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a thin circular wire of radius r has a charge Q. if a point charge q is placed at the centre of the ring, then find the increase in the tension of the wire.
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2 Jul 2007 12:49:42 IST
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Lets take a small part of ring in the form of arc with length dl which suspends angle  on the centre of the ring.
Force b/w charge q and arc with charge=dq= (q/2 r)*dl, df= q*dq/r2
this force df=2*Tsin /2. As is very small so replace sin/2 to /2.
df= 2T/2
df=T (=dl/r)
T=df/
T=(q*q*dl*r)/dl*r*r*2r
T=q2/2pie*r*r
in above solution i have not taken dielectric constant ..... try to draw the diagram...
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Good work maverick
Before the charge was placed there was no tension. So the increment in tension is the value of tension after the charge is placed
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2 Jul 2007 13:53:23 IST
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the force stretching the wire will be the tension acting tangentially.if  is the angle between the tangential components , resolve the tension and integrate to find the force. 2dtsin /2= [ 0] [ 2pir] kdQq/r*r 2tsin /2= [ 0][ 2pir] kdxQq/2 pi r*r*r (since dQ=dxQ/2 pi *r) on integrating u'll have -----t=Qq/8 pi  0 r*r |
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