two charges 20×10^-6C and 1.0×10^-6 are placed at a sepration of 10 cm . where should a third charge be placed such that it experiences no net force due to these charges?
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if both are positive charges then the third charge should be placed on the line joining the two charges.
it should be in between the two charges
F1=F2
2*q/ r2=1*q/(10-r)2
r2=2(10-r)2
r2= 200 + 2 r2 - 40r
r2 - 40r + 200 = 0
r = [40 ± √(1600-800)]/2
= [40 - √(1600-800)]/2 since the point should be in between
= 20 - 10√2
= 5.86 cm
third charge should be placed at 5.86 cm from 2*10-6 C charge.
it should be in between the two charges
F1=F2
2*q/ r2=1*q/(10-r)2
r2=2(10-r)2
r2= 200 + 2 r2 - 40r
r2 - 40r + 200 = 0
r = [40 ± √(1600-800)]/2
= [40 - √(1600-800)]/2 since the point should be in between
= 20 - 10√2
= 5.86 cm
third charge should be placed at 5.86 cm from 2*10-6 C charge.
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