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(1) What is the excess number of electrons that must be placed on each of two small sphere spaced 3cm apart so that a force of repulsion that acts between them becomes 10-19N

(a)25

(b)225

(c)625

(d)125

Please show the solution !!!!!!!!!!!!

Nimish Singh , 12 Years ago
Grade 11
anser 4 Answers
Abhishekh kumar sharma

225

Last Activity: 12 Years ago
saurabh gupta

here q1=q2

10-19=(9*109)(q1.q2)/32

10-28=q1q2

10-14C=q1

 

q=ne

10-14=n*1.6*10-19

n=625

Last Activity: 12 Years ago
BAYANA SAGAR

I think that (c)is the correct answer.because no of electrons is 6.25( -18).10  

Last Activity: 12 Years ago
Mahendra Chapagain
625
solution:
we know F=K.qq/r^2
applying in above question F=K.(ne)^2/r^2
Now N^2=F.r^2/(e^2.K) where k is constant value 8.98*10^9
finally by solving we get n=625
 
Last Activity: 7 Years ago
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