Question

A rod AB of length $L$ and mass $m$ is uniformly charged with a charge Q and it is suspended from end A as shown in the figure. The rod can freely rotate about A in the plane of figure. An electric field $E$ is suddenly switched on in the horizontal direction due to which rod gets turned by a maximum angle ${90}^o$. The magnitude of E is equal to $\frac{nmg}{Q}$. Find the value of n.

Answer 1

Since the rod is suspended freely initially it means its kinetic energy initially is 0.
Thereafter that it gets turned by ${90}^o$ but at this point also its velocity is 0 which means its kinetic energy at this point is also 0.
$\Delta KE=0$
Consider the point where center of mass of the rod lies (in vertical position) as(0,0).
Let us now consider all the forces acting on the rod while it is being rotated by ${90}^o$.
When the rod is getting rotated by ${90}^o$ its centre of mass also gets shifted and some work will be done by both gravitational force($W_g$) and electric field($W_E$).

By work energy theoream we know
$\Delta K.E=W_g+W_E$
$⟹0=-mg\left(\frac{l}{2}\right)+QE\frac{l}{2}$
$\therefore E=\frac{mg}{Q}$