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when did a charge become zero at an electric field having charges 16microcoulomb and -9mc.they have a distance of 8cm between them

dia , 8 Years ago
Grade 12
anser 1 Answers
Vikas TU
As we realize that 
 
E.F. = kq/r^2 
 
q = charge , r = separate 
 
Electric field will be zero somewhere close to them 
 
That is 
 
Net E.f. = k(16 x10^-6)/r^2 + k(9x10^-6)/(8-r)^2 
 
0 = k(16 x10^-6)/r^2 + k(9x10^-6)/(8-r)^2 
 
k(9x10^-6)/(8-r)^2 = k(16 x10^-6)/r^2 
 
=> 9/(8-r)^2 = 16/r^2 
 
=> 16(64+r^2-16r) = 9r^2 
 
From this we get,
 
=> r = 5.12 
Last Activity: 8 Years ago
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