91Ó°ÊÓ

1. Along non-conducting rod of mass less rod of length L is balanced by weight W, and at the left and right ends is hanging two conducting spheres of charges q and 2q.
2. At a distance h, under this system is a sphere of charge Q.

Using the concept of Coulomb's law, we can get the force acting on the system. To get the equation of rotational equilibrium, we take the values of all torques to be zero. This determines the required values of x and h.

Formulae:

The magnitude of the electrostatic force between any two particles,$F=k\frac{q_1{q}_2}{r^2}$ (i)

The torque acting on a body can be given as:$\mathrm{τ}=F×r$ (ii)

The charge Q on the left exerts an upward force which is given by using equation (i) as:(at a distance L/2 from the bearing)

$F_1=\frac{Qq}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{h}^2}$

Now, the torque acting on the force is given by using equation (ii) as:

$\begin{array}{l}{\mathrm{Ï„}}_1=\frac{−Qq}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{h}^2}\left(\frac{L}{2}\right)\mathrm{............................}\text{(a)}\\ \text{(asforceisoppositetothedirectionofbearing)}\end{array}$

Now, the attached weight exerts a downward force of magnitude,$F_3=W$, at a distance (x - L /2) from the bearing.

Thus, the torque for this case is given using equation (ii) as:

$\begin{array}{l}{\mathrm{τ}}_2=−W\left(x−\frac{L}{2}\right)\mathrm{............................}\text{(b)}\\ \text{(asforceisoppositetothedirectionofbearing)}\end{array}$

Now, the charge Q on the right exerts an upward force which is given using equation (i) as: (at a distance L/2 from the bearing.)

$F_3=\frac{2Qq}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{d}^2}$

Now, the torque for this case is given using equation (ii) as:

$\begin{array}{l}{\mathrm{Ï„}}_3=\frac{2Qq}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{d}^2}\left(\frac{L}{2}\right)\mathrm{..............................}\left(c\right)\\ \left(\text{astheforceisactinginthedirectionofthebearing}\right)\end{array}$

Since the rod is in equilibrium, the net force acting on it is zero, and the net torque about any point is also zero.

Hence, the equation of rotational equilibrium using (a), (b) and (c) is given as: $\begin{array}{c}\frac{−1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{qQ}{h^2}\frac{L}{2}−W\left(x−\frac{L}{2}\right)+\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2qQ}{h^2}\frac{L}{2}=0\\ x=\frac{L}{2}\left(1+\frac{qQ}{W{h}^2}\right)\end{array}$

Hence, the value of x is$\frac{L}{2}\left(1+\frac{qQ}{W{h}^2}\right)$

If $F_N$is the magnitude of the upward force exerted by the bearing, then using Newton’s second law (with zero acceleration) $F_N=0$,gives us the above rotational equilibrium equation as:

$\begin{array}{c}\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{qQ}{h^2}−\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2qQ}{h^2}−F_N=0\\ h=\sqrt{\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{qQ}{W{h}^2}}\end{array}$

Hence, the value of h is$\sqrt{\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{qQ}{W{h}^2}}$