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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 May 2007 14:03:16 IST
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plz help me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Q.- A partical 'A' having a charge of 2 micro Coloumb is held fixed on a horizontal
table.A second charged partical of mass 80g stays in equilibirium on the table at a distance of 10cm from the first charge.
The coefficient of friction between the table and this second partical is u=0.2
Find the range within which the charge of this second partical may lie.
give solution quick
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hi i am Abhay. :)
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Shoot for the moon............ if u missed........ still u will be amoung the stars.!!
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17 May 2007 14:26:59 IST
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i'm neglecting the gravitational force between the two bodies here
frictional force = 0.2*0.08 = 0.016 N
force due to electrostatic interaction = 9 * 10^9 * 2 * 10^(-6) * q/ 0.01
when f due to electostatic interaction < fric force , the body b will be in equilibrium
0.016 = 18 * 10^5 * q
|q| < 0.88 * 10^(-8)
q can lie in the range [ - 8/9 * 10^(-8) ; 8/9 * 10^(-8) ; ]
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