Question

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance is maximum?

Answer 1

`q / q_(1) = 2`, assuming that
`q_(1)` and
`(q - q_(1))` are point charges.

Step-by-step explanation:

Let
`k` denote the coulomb constant. Let
`r` denote the distance between the two point charges. In this question, neither
`k` and
`r` depend on the value of
`q_(1)`.

By Coulomb's Law, the magnitude of electrostatic force between
`q_(1)` and
`(q - q_(1))` would be:

`\begin{aligned}F &= (k\, q_(1)\, (q - q_(1)))/(r^(2)) \\ &= (k)/(r^(2))\, (q\, q_(1) - {q_(1)}^(2))\end{aligned}`.

Find the first and second derivative of
`F` with respect to
`q_(1)`. (Note that
`0 < q_(1) < q`.)

First derivative:

`\begin{aligned}(d)/(d q_(1))[F] &= (d)/(d q_(1)) \left[(k)/(r^(2))\, (q\, q_(1) - {q_(1)}^(2))\right] \\ &= (k)/(r^(2))\, \left[(d)/(d q_(1)) [q\, q_(1)] - (d)/(d q_(1))[{q_(1)}^(2)]\right]\\ &= (k)/(r^(2))\, (q - 2\, q_(1))\end{aligned}`.

Second derivative:

`\begin{aligned}\frac{d^(2)}{{d q_(1)}^(2)}[F] &= (d)/(d q_(1)) \left[(k)/(r^(2))\, (q - 2\, q_(1))\right] \\ &= ((-2)\, k)/(r^(2))\end{aligned}`.

The value of the coulomb constant
`k` is greater than
`0`. Thus, the value of the second derivative of
`F` with respect to
`q_(1)` would be negative for all real
`r`.
`F\!` would be convex over all
`q_(1)`.

By the convexity of
`\! F` with respect to
`\! q_(1) \!`, there would be a unique
`q_(1)` that globally maximizes
`F`. The first derivative of
`F\!` with respect to
`q_(1)\!` should be
`0` for that particular
`\! q_(1)`. In other words:

`\displaystyle (k)/(r^(2))\, (q - 2\, q_(1)) = 0`.

`2\, q_(1) = q`.

`q_(1) = q / 2`.

In other words, the force between the two point charges would be maximized when the charge is evenly split:

`\begin{aligned} (q)/(q_(1)) &= (q)/(q / 2) = 2\end{aligned}`.