• The two point charges +9e  and +e  are placed at a distance of 16cm from  each other .At what point between the charges .At what point between the charges should the third charge q be placed so that it remains in equilbirium?
  • Three point charges q1 ,q2,q3 are in line at equal distances q2 and q3 are in sign .Find the magnitude and sign of q1 ,if the net force on q3 is zero.
  • ABC is an equilateral triangle of side 10 m and D is the midpoint of BC. charges of +100 miocrocoulomb ,-100 microcoulomb and +75microcoulomb are placed at B C and D respectively .Find the force on a +microcoulomb placed at A.
  • At each of the four corners of a square  of side a ,a charge +q is placed  freely .What charge should be placed at the centre so that the whole system be in equilbirium?


1st  case

Given charges: +e and +9e

Let the third chare q be placed at a distance x from e and hence its distance from 9e is 16-x cm

For the system to remain in equilibrium,

F1 = F2


F1 = K(e q)/x2

F2 = K(9e)/(16-x)2

Equating F1 and F2 , we get,

9/x2 = 1/(16-x)2

Taking square root on both sides:

3/x = 1/(16-x)

Solving for x, we get,

x = 12cm

hence the charge q should be placed at 12 cm from e.

2nd case

See the figure below,

Now net force at q3 is 0 as given in the question. Hence,

Applying coloumb’s law on q3 we get,

q3q1K/(2x)2 + Kq2q3/x2 = 0

q1/4 +q2 = 0

q1  = - q2/4

3rd case

F1 between charge placed at A and charge placed at B :

= K*100*1*10-12/100

= K*10-12N

F2 between D and A

= K*75*1*10-12/AD2

Now AD = 75

Hence F2 = K*75*10-12/75

F2 = K*10-12N

Resultant R between F1 and F2 is (F12+F22+2F1F2cos300)

= (2F12 +2F12*3/2)

= F1(2+3) = K*10-12√(2+√3)

Now force F3 between C and A is :


= -K*100*10-12/100

= -K*10-12N

The resultant of these two forces will be the net force on charge placed at A.

That is resultant of R and F3

R1 = √(R2 +F32 +2RF3cos135)

R1 = √((K*10-12√(2+√3))2 + (-K*10-12)2 -2(K*10-12)2√(2+√3)(-1/√2))

R1 = K*10-12√(2+√3 +1+√2√(2+√3))

R1 = 9*109*10-12√(3 +√3 +√(4 +2√3))N . this is the required force.

Case 4

See the figure below:

Here since the system is in equilibrium hence the net force on each charge q will be 0. We are considering 1 such charge. Here:

Force F = Kq2/a2 +Kq2/a2 + the forces due to diagonal element

First lets find he resultant of the 1st two forces:

R = √2(Kq2/a2)

Now force due to diagonal  charge q :

F1 = Kq2/(a√2)2 where a√2 is the length of the diagonal

F1 = Kq2/2a2

Force F2 due to Q :

F2 = KqQ/(a√2/2)2

Net force on q hence = R +F1+F2 = 0

√2Kq2/a2 +Kq2/2a2 +2KqQ/a2 = 0

Cancelling Kq/a2 on both sides:

Q = -(4√2 +2)q/8


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