force on elementary length. = Eλdx
torque=Eλdxxsinθ
integrating from 0 to l we get λEsinθl2/2
similarly for lower part (-λ)
torque in same sense. λEsinθl2/2
so net torque = λEsinθl2
E=σ/2ε0
so τ= λEsinθl2/2ε0
a large sheet carries uniform surface charge density sigma and a rod of length 2l has a linear charge density lambda on 1 half and -lambda on other half the rod is hinged at midpoint O and makes an angle theta with the normal to the sheet....the torque experienced by the rod is....
force on elementary length. = Eλdx
torque=Eλdxxsinθ
integrating from 0 to l we get λEsinθl2/2
similarly for lower part (-λ)
torque in same sense. λEsinθl2/2
so net torque = λEsinθl2
E=σ/2ε0
so τ= λEsinθl2/2ε0
as force in constant so u could have used (λl)Esinθ(l/2) for torque on upper part ...(force on center of rod. at l/2)
similarly for lower part.