CBSE Class 12-science Answered
At what separation should two equal charges , 1.0 C each, be placed so that the force between them equals the weight of a 50 kg person
Asked by haroonrashidgkp | 20 Apr, 2018, 11:26: AM
Expert Answer
Force due to charges F = (q1×q2)/ [ (4πε0) r2 ] ; q1 and q2 are charges in Coloumb. is permisibility in air/vacuum. r is the separation distance
we are given F = 50 kgf = 50×9.8 N ; q1 = q2 = 1C ;
1 / (4πε0) = 8.89×109 ;
hence 50×9.8 = 8.89×109 / r2 ;
r2 = 8.89×109/(50×9.8) ; solving for r we get r = 4284 m
Answered by Thiyagarajan K | 20 Apr, 2018, 02:37: PM
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