14. Two point-charges +9e and te are placed at a 2distance of 16 cm fr

1.	14. Two point-charges +9e and te are placed at a 2distance of 16 cm from each other. At what pointbetween these charges should a third charge q beplaced so that it remains in equilibrium?Ans. At a distance of 12 cm from the charge +9e.
Answer» 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 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. 2ndcase 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 hit like if you find it useful

14. Two point-charges +9e and te are placed at a 2distance of 16 cm from each other. At what pointbetween these charges should a third charge q beplaced so that it remains in equilibrium?Ans. At a distance of 12 cm from the charge +9e.

Answer»

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

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.

2ndcase

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

hit like if you find it useful

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