Question

N fundamental charges each of charge ‘q’ are to be distributed as two point charges separated by a fixed distance, then calculate the ratio of maximum to minimum force. (N is even and greater than 2)

Answer 1

Hint: To solve this problem we must have the knowledge of the formula for the electrostatic force between two point charges. We also required the relation between arithmetic mean and geometric mean of two natural numbers.

Formula used:
$F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

Complete step by step answer:
It is given that there are N number of charges with charge q (each). Therefore, the net charge of the system is ‘Nq’. Let the two point charges be ‘$xq$’ and $(N-x)q$, where x is a natural number less than N. The distance of separation of the two point charges be r.The electrostatic force between the two charges (say ${{q}_{1}}$ and ${{q}_{2}}$) separated by a distance r is given as $F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$, where k is a proportionality constant.
In this case, ${{q}_{1}}=xq$ and ${{q}_{2}}=(N-x)q$.
$\Rightarrow F=\dfrac{k(xq)(N-x)q}{{{r}^{2}}}$
$\Rightarrow F=\dfrac{k{{q}^{2}}(x)(N-x)}{{{r}^{2}}}$ …. (i).
From (i) we can understand that the magnitude of the force depends on the way that the fundamentals charges are distributed, i.e. it depends on the value of x.

Consider the term $(x)(N-x)$.Here, numbers in both the brackets are natural numbers. Therefore, we can write that $AM\ge GM$, where AM is the arithmetic mean and GM is the geometric mean of the two numbers respectively.
Here, $AM=\dfrac{x+(N-x)}{2}=\dfrac{N}{2}$
And $GM=\sqrt{x(N-x)}$.
$\Rightarrow \dfrac{N}{2}\ge \sqrt{x(N-x)}$.
$\Rightarrow \dfrac{{{N}^{2}}}{4}\ge x(N-x)$
This means that the maximum value of $x(N-x)$ is equal to $\dfrac{{{N}^{2}}}{4}$.
$\Rightarrow \dfrac{{{N}^{2}}}{4}=Nx-{{x}^{2}}$
$\Rightarrow {{x}^{2}}-Nx+\dfrac{{{N}^{2}}}{4}=0$.
But ${{x}^{2}}-Nx+\dfrac{{{N}^{2}}}{4}={{\left( x-\dfrac{N}{2} \right)}^{2}}$
$\Rightarrow {{\left( x-\dfrac{N}{2} \right)}^{2}}=0$
This means that $x-\dfrac{N}{2}=0$.
$\Rightarrow x=\dfrac{N}{2}$.
Therefore, the maximum value of F is when $x=\dfrac{N}{2}$.

Substitute $x=\dfrac{N}{2}$ in (i).
${{F}_{max}}=\dfrac{k{{q}^{2}}\left( \dfrac{N}{2} \right)\left( N-\dfrac{N}{2} \right)}{{{r}^{2}}}=\dfrac{k{{q}^{2}}{{N}^{2}}}{4{{r}^{2}}}$ … (ii).
The minimum value of F will be when $x=1$.
Substitute $x=1$ in (i).
$ {{F}_{\min }}=\dfrac{k{{q}^{2}}\left( 1 \right)\left( N-1 \right)}{{{r}^{2}}}\\
\Rightarrow {{F}_{\min }}=\dfrac{k{{q}^{2}}\left( N-1 \right)}{{{r}^{2}}}$ …. (iii).
Now, divide (ii) by (iii).
$\dfrac{{{F}_{max}}}{{{F}_{\min}}}=\dfrac{\dfrac{k{{q}^{2}}{{N}^{2}}}{4{{r}^{2}}}}{\dfrac{k{{q}^{2}}(N-1)}{{{r}^{2}}}}\\
\therefore \dfrac{{{F}_{max}}}{{{F}_{\min }}}=\dfrac{{{N}^{2}}}{4(N-1)}$.

This means that the ratio of the maximum force to the minimum force is $\dfrac{{{N}^{2}}}{4(N-1)}$.

Note: We can also find the maximum value of F by differentiation.If we differentiate the F with respect to x and equate it to zero, then we will obtain the value of x for which F is maximum. We can write the equation (i) as,
$F=\dfrac{k{{q}^{2}}\left( -{{x}^{2}}+Nx \right)}{{{r}^{2}}}$.
This equation is the equation of a downward parabola. In this, the value of F first increases, reaches a maximum value then it begins to decrease. Therefore, the minimum value of F is for $x=1$.