• 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 ,q_2,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?

 Hi,

1^st  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

Therefore,

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.

2^nd 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

3^rd 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*1*10^-12/AC²

= -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 1^st 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