A large sheet carries uniform surface charge density sigma.A rod of length 2l has a linear charge density lamda on one half and -lama on the other.The rod is hinged at mid point O and makes theta with normal to the sheet.The torque experienced by the rod is
Hi,
Elementary length. = Eλdx
Now, torque = Eλdxxsinθ
Elementary length. = Eλdx
Now, torque = Eλdxxsinθ
Thus, we get λEsinθl2/2
similarly for lower part torque will be λEsinθl2/2
so net torque = λEsinθl2
E=σ/2ε0
so τ = λEsinθl2/2ε0
