In the figure shown S is a large nonconducting sheet of uniform charge density `sigma`. A rod R of length `l` and mass 'm' is parallel to the sheet and hinged at its mid point. The linear charge densities on the upper and lower half of the rod are shown in the figure. Find the angular acceleration of the rod just after it is released.

Answer

`(3sigma lambda)/(2m in_(0))`

Explanation

Net force on upper half`=((lambdal)/2)(sigma/(2in_(0)))`
Torque on upper half`=((lambdal)/2)xxsigma/(2in_(0))xxl/4`
[`l/4=` distance from hinge point for centre of mass]
total torque on rod
`=(lambdal)/2xxsigma/(2 in_(0))xxl/4xx2=I alpha=(ml^(2))/12 prop :. alpha=(3lambdasigma)/(2m in_(0))`