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- 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 q
_{1 },q_{2,}q_{3} are in line at equal distances q_{2 }and q_{3} are in sign .Find the magnitude and sign of q_{1 },if the net force on q_{3} 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?

_{1 },q_{2,}q_{3}are in line at equal distances q_{2 }and q_{3}are in sign .Find the magnitude and sign of q_{1 },if the net force on q_{3}is zero.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)/x^{2}

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

Equating F1 and F2 , we get,

9/x^{2} = 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/x^{2} = 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^{-12}N

F2 between D and A

= K*75*1*10^{-12}/AD^{2}

Now AD = √75

Hence F2 = K*75*10^{-12}/75

F2 = K*10^{-12}N

Resultant R between F1 and F2 is √(F1^{2}+F2^{2}+2F1F2cos30^{0})

= √(2F1^{2} +2F1^{2}*√3/2)

= F1√(2+√3) = K*10^{-12}√(2+√3)

Now force F3 between C and A is :

-K*100*1*10^{-12}/AC^{2}

= -K*100*10^{-12}/100

= -K*10^{-12}N

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

That is resultant of R and F3

R1 = √(R^{2} +F3^{2} +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*10^{9}*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 = Kq^{2}/a^{2} +Kq^{2}/a^{2} + the forces due to diagonal element

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

R = √2(Kq^{2}/a^{2})

Now force due to diagonal charge q :

F1 = Kq^{2}/(a√2)^{2} where a√2 is the length of the diagonal

F1 = Kq^{2}/2a^{2}

Force F2 due to Q :

F2 = KqQ/(a√2/2)^{2}

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

√2Kq^{2}/a^{2} +Kq^{2}/2a^{2} +2KqQ/a^{2} = 0

Cancelling Kq/a^{2} on both sides:

Q = -(4√2 +2)q/8

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