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Two charges2*10^-6c and 1*10^-6c are placed at a seperation of 10 cm.Where should a third charge be placed such that it experiences no net force due to these charges?

```
11 years ago

```							Hi Pallavi,
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.
```
11 years ago
```							let us assume that the charge (Q) is at ‘r’ distance from charge 2*10-6 C hence it is (10-r) distance from another charge i.e 1*10-6 C. Now using coulomb’s formula force applied on charge Q by Charge 2*10-6 C is F=kq1Q/r2...Eq1 And the force applied on charge Q by charge 1*10-6 is F=kq2Q/(10-r)2.....Eq 2As the net force should be zero therefore Eq 1- Eq 2=0 kq1Q/r2 – kq2Q/(10-r)2 =0kq1Q/r2 = kq2Q/(10-r)2 k and Q gets cancelled, we are left withq1/r2 =q2/(10-r)2   ..... Eq 3  we know that (q1=2*10-6 C & q2 = 1*10-6 C)put the values in eq 32*10-6 /r2  = 1*10-6 /(10-r)2cross multiply the equationr2/(10-r)2  =  2*10-6 / 1*10-6 r2/(10-r)2  =  2 / 1either make a quadratic equation and then solve or Take sq root on both sides, i am taking square roots on both the sides as it is easier for me to solver/(10-r)  = under root 2r = 10 root 2 – r root 2r+r root 2 = 10 root 2take r commonr =10 root 2/ (1+root 2)   (we all know root 2 =1.4)on solving u will get 5.85 cm as answer
```
2 years ago Yash Chourasiya
246 Points
```							Hello StudentLet us assume that the charge q3is placed at a distance of 'x' from q1and we take the distance between q1and q2as 10 cm (0.1m).In order for the third charge (q3) to be in equilibrium, the forces exerted on it by charge q1and q2should balance outthus,kq3q1/x2- kq3q2/(10-x)2= 0 …..(1)whereq1= 2X10-6Cq2= 10-6Cand we have to calculate x.now, further solving (1), we getq2/ (10 - x)2= q1/x2or x = 10 / [1 + (q1+ q2)1/2]by putting the values of q2and q1, we getx = 10 / 1.706or x = 5.86 cmso, the charge should be placed at a distance of 5.86 cm from the charge q1or 4.14 cm from charge q2.I hope this answer will help you.
```
4 months ago
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