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Two charges210^-6c and 110^-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?

pallavi pradeep bhardwaj , 16 Years ago

Grade 12

3 Answers

askiitian.expert- chandra sekhar

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.

Last Activity: 16 Years ago

nirupama singha

let us assume that the charge (Q) is at ‘r’ distance from charge 2*10^-6C hence it is (10-r) distance from another charge i.e 1*10^-6C.

Now using coulomb’s formula force applied on charge Q by Charge 2*10^-6C is F=kq₁Q/r²...Eq1

And the force applied on charge Q by charge 1*10^-6is 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/r^{2 =}q₂/(10-r)²..... Eq 3

we know that (q₁=2*10^-6C & q₂= 1*10^-6C)

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

Last Activity: 7 Years ago

Yash Chourasiya

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.

Last Activity: 5 Years ago