Please look Q$($a$)$ thoroughly!!!!!Do not DEVIATE on the question!!!!!!!

The diagram shows two spheres of mass $m$ suspended from wires of negligible

mass. The length between the pivot points and the centres of each sphere is $L.$ The

two suspension points lie on a horizontal line and are separated by a distance $x.$ The

system is in equilibrium. Take the total acceleration due to gravity to be $g.$

Hint: You may assume that $x$&$L$ as well as $x$&$x.$

$($a$)$ Show that the small deflection $x$ by which the gravitational force $F_9$ between the

spheres will deflect them from the vertical is given by:

$x$~~$\frac{x}{8}-\sqrt[\phantom{2}]{(\frac{x}{8}{)}^2-\frac{LmG}{4xg}}$

where G is Newton's gravitational constant. Clearly state all assumptions you

make and show all your workings.

$($b$)$ The spheres are now both electrically charged with the same positive charge $Q.$

How big must $Q$ be so that the wires hang vertically again?

Hint: The force between two spheres, electrically charged with charges $Q_1$ and $Q_2$

is given by:

$F_Q=k\frac{Q_1{Q}_2}{r^2}$

where $k$ is a constant and $r$ is the separation of their centres.