Two charges, each of `+10^(3)` esu, are placed at two points A and B separated by a distance of 200 cm. From the middle point of AB, along its perpendicular bisector, a particle having `-10^(3)` esu of charge is thrown upwards with energy `10^(4)` erg. Determine the maximum height attained by the particle. The effect of gravitation can be neglected.

Explanation

Let us assume that the charged particle turns back from D.

Let the distance between C and D be x cm. for moving the charged particle from C to D, we may write from the conservation of mechanical energy, initial kinetic energy = increase in electrical potential energy.
Potential at C due to charge placed at A and B,
`V_(C ) =1000/(AC) + 1000/(BC)`
But `AC = BC = 100 cm`
`:. V_(C) = 1000/100 + 1000/100 = 20 esu`
When -1000 esu of charge is at C then potential energy of the system,
`U_(C) = V_(C) xx (-1000) = -20 xx 10^(3) erg`
Again, potential at D due to charge placed at A and B,
`V_(D) = 2 xx (1000)/(sqrt(x^(2)+100^(2)))esu`
Thus, when -1000 esu of charge is at D, potential energy of the system, `U_(D) = V_(D) xx (-1000) = 2 xx (1000)/(sqrt(x^(2)+100^(2))) (-1000) erg`
Accordingly,
`U_(D) - U_(C)` = initial kinetic energy of the particle at C
or, `[20-(2xx10^(3))/(sqrt(x^(2)+10^(4)))] xx 10^(3) = 10^(4)`
`or, (-2 xx 10^(3))/(sqrt(x^(2)+10^(4))) = 10 -20`
`or, sqrt(x^(2) + 10^(4)) = 200`
`:. x = 173.2 cm`.
Therefore, the particle will turn back after covering a distance of 173.2 cm.