A brilliant year

Sorry for a mistake in the problem previously! I have fixed it now!

Now the system says that $73$ strings are attached to a point which are attached to balls. We will assume that the first string is on the y-axis and subsequent strings a definite angle apart. Also, here we have just used $\epsilon$ for ${\epsilon}_{0}$ just to ease our writing(though it is wrong as $\epsilon = \kappa{\epsilon}_{0}$ Now, to be in equilibrium,

Tension in the strings would be equal to weight-force . Now this would be the case for all $73$ balls. Hence,

Y-component is

$\displaystyle 73\text{mg} -({T}_{1} + {T}_{1}cos\theta + {T}_{1}cos2\theta + {T}_{1}cos3\theta + \cdots + {T}_{1}cos72\theta)$ [as all strings are identical, all have the same tensions]

Also,

X -component is

$\displaystyle {T}_{1} + {T}_{1}sin\theta + {T}_{1}sin2\theta + \cdots + {T}_{1}sin72\theta$

Now, y-component is zero but balls have a charge as well. So, they will have an electrostatic force as well and that will be along the x-axis.

The distance between 2 balls is given by a cosine rule result.

$\displaystyle \text{Distance between rth and kth ball} = 2{l}^{2}(1 - cos((k-r)\theta))$

Therefore, the sum of electrostatic forces between balls will be

$\displaystyle \sum_{i=1}^{72}{\sum_{k=1}^{i}{\frac{{q}^{2}}{8\pi\epsilon{l}^{2}(1-cos(k\theta))}}}$ [This is because each ball will have a force of interaction with every other ball, and the force can be considered as coulomb force]

Now, the x-components will be:

$\displaystyle {T}_{1}(\sum_{k=0}^{72}{sink\theta}) = \frac{{q}^{2}}{8\pi\epsilon{l}^{2}}\displaystyle \sum_{i = 1}^{72}{\frac{72-i}{1-cosi\theta}}$

Now, y-component will be:

$\displaystyle {T}_{1}(\sum_{m=0}^{72}{cosm\theta}) = 72mg$

$\displaystyle {T}_{1} = \frac{72mg}{\displaystyle \sum_{m=0}^{72}{cosm\theta}}$

Now substituting this in equation of x-axis,

$\displaystyle \frac{72mg}{\displaystyle \sum_{m=0}^{72}{cosm\theta}}(\sum_{k=0}^{72}{sink\theta}) = \frac{{q}^{2}}{8\pi\epsilon{l}^{2}}\displaystyle \sum_{i = 1}^{72}{\frac{72-i}{1-cosi\theta}}$

After some bashing we get,

$\displaystyle m = \frac{{q}^{2}\displaystyle \sum_{i = 1}^{72}{\frac{72-i}{1-cosi\theta}}cot(36\theta)}{72*8\pi\epsilon{l}^{2}g}$

Well, how we got that $cot$ is by using the formulas:

$\displaystyle sin(a) + sin(a+b) + sin(a+2b) +\cdots+ sin(a+(n-1)b) = \frac{sin(nb/2)}{sin(b/2)}(sin(a+(n-1)b/2))$

and

$\displaystyle cos(a) + cos(a+b) + cos(a+2b) +\cdots+cos(a+(n-1)b) = \frac{sin(nb/2)}{sin(b/2)}(cos(a+(n-1)b/2))$

Now, just we have to do is to substitute some values in the expression -

$\displaystyle \theta = 1, \text{q} = 2, \text{l} = 17$

$\displaystyle m = \frac{{2}^{2}\displaystyle \sum_{i = 1}^{72}{\frac{72-i}{1-cosi\theta}}cot(36)}{576\pi\epsilon{17}^{2}g}$

The sum is difficult to handle and I am still finding problem to find its analytic answer and so you can use Wolfram Alpha here.

As a result,

$\displaystyle m = \frac{364.9245675}{576g\pi\epsilon}$

The denominator is just $\displaystyle 1.57E-7$ and that's what we have to multiply to $m$ .

Therefore, $a$ is

$\displaystyle 364.9$ and here you get the answer as $\displaystyle \boxed{365}$