A rod AB of mass M and length L is positively charged with liner charge density 1 col/m.It is suspended at end A and is hanging freely.If a horizontal electric field E is switched on,find the angular acceleration

moment of inertia =I =ML²/3, assume that the linear charge density is constant. Torque=(L/2)(kLE),k is linear charge density.

Torque=I*(alpha), where 'alpha' is angular acceleration.After simplifying we get α=(3kE/2M)