Three point charges q , 2q and 8q are to be placed on a 0.09m long straight line. Find the positions of the charges so that the potential energy of this system becomes minimum. In this situation, find the intensity at the position of the charge q due to the other two charges?

Explanation

Let the charge 2q be placed in between the charges q and 8q and the distance between q and 2q be x metre.
So, the distance between `2q and 8q = (0.09 -x)m`

`:.` Potential energy of the system,
`U = 1/(4 pi epsilon_(0)) [(qcdot 2q)/x + (2q cdot8q)/((0.09-x)) + (q cdot 8q)/(0.09)]`
`= 9 xx 10^(9) xx 2q^(2) [1/x + 8/((0.09-x))+4/0.09]`
For minimum value of `U, (dU)/(dx) = 0`
i.e., `9 xx 10^(9) xx 2q^(2)[-1/(x^2)+8/((0.09-x)^(2))]=0`
or, `1/x = (2sqrt2)/(0.09-x) = (2.83)/(0.09-x) or, x = 0.0235`
So the charge 2q is to be placed at a distance of 0.0235 m, i.e., 2.35 cm from the charge q.
Electric field intensity at the position of the charge q due to the other charges,
`E = 9 xx 10^(9) xx (2q)/(x^2) + 9 xx 10^(9) xx (8q)/((0.09)^(2))`
`= 9 xx 10^(9) xx 2q[1/((0.0235)^(2))+4/((0.09)^(2))]`
`=4.148 xx 10^(13) q N cdot C^(-1)`.