Question

two point charge +9e and +e units are placed at a distance r apart .find the pisition of third point charge so that it is in equilibrium​

Answer 1

Answer:

3r/4

Explanation:

+9e____x_______(q)_______(r-x)___+e

the above diagram well explains the situation

r is the total distance as given

and x is our assumption of the distance of charge 9e to q

and thus the distance between q and e will be r-x

now for equilibrium, the vector sum of the two charges should be equal to 0

F(net)=0

F1 = K(e q)/x2

F2 = K(9e q)/(r-x)2

Equating F1 and F2 , we get,

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

Taking square root on both sides:

3/x = 1/(r-x)

Solving for x, we get,

3r-3x=x

3r=4x

x=3r/4