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HETAV PATEL

Grade 11,

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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?

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?

Grade:12

3 Answers

askiitian.expert- chandra sekhar
10 Points
14 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.

nirupama singha
15 Points
5 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 2
As the net force should be zero therefore Eq 1- Eq 2=0
 kq1Q/r2 – kq2Q/(10-r)2 =0
kq1Q/r2 = kq2Q/(10-r)2 
k and Q gets cancelled, we are left with
q1/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 3
2*10-6 /r2  = 1*10-6 /(10-r)2
cross multiply the equation
r2/(10-r)2  =  2*10-6 / 1*10-6 
r2/(10-r)2  =  2 / 1
either 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 solve
r/(10-r)  = under root 2
r = 10 root 2 – r root 2
r+r root 2 = 10 root 2
take r common
r =10 root 2/ (1+root 2)   (we all know root 2 =1.4)
on solving u will get 5.85 cm as answer
 
 
 
Yash Chourasiya
askIITians Faculty 256 Points
3 years ago
Hello Student

Let 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 out

thus,

kq3q1/x2- kq3q2/(10-x)2= 0 …..(1)

where
q1= 2X10-6C
q2= 10-6C
and we have to calculate x.

now, further solving (1), we get

q2/ (10 - x)2= q1/x2

or x = 10 / [1 + (q1+ q2)1/2]

by putting the values of q2and q1, we get

x = 10 / 1.706

or x = 5.86 cm

so, 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.

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