badge image

Enroll For Free Now & Improve Your Performance.

User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12


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

Answers : (3) chandra sekhar
10 Points

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


2*q/ r2=1*q/(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
nirupama singha
15 Points
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 & q= 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*10-6 / 1*10-6 
r2/(10-r) =  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
2 years ago
Yash Chourasiya
askIITians Faculty
246 Points
							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


kq3q1/x2- kq3q2/(10-x)2= 0 …..(1)

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.
4 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details