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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 -19 N (a)25 (b)225 (c)625 (d)125 Please show the solution !!!!!!!!!!!!

(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 !!!!!!!!!!!!

Grade:11

4 Answers

Abhishekh kumar sharma
34 Points
10 years ago

225

saurabh gupta
14 Points
10 years ago

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

BAYANA SAGAR
48 Points
10 years ago

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

Mahendra Chapagain
13 Points
5 years ago
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
 

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