Recent content by microdosemishief

I just read the textbook answer but I want to know why my answer is wrong and not just a different method to the same answer

R

Yes and dE = kdQ/r^2 so the denominator is r^2 where r = Rcos(θ) squared, because dE_x is Rcosθ distance away

I wrote that in my question

My attempt: due to symmetry along x-axis, net E is only along x^hat. dQ = λ dl = λ (R dθ) for each dl, the x component of distance from dl to the origin is Rcos(θ) Hence, E_x = \int_{-θ_0}^{θ_0} k(λ R dθ)/(Rcos(θ)^2 = kλ/R \int_{-θ_0}^{θ_0} sec(θ)^2 dθ = 2kλ/R tan(θ_0) along negative x^hat...

Assuming initial displacement is zero, if x = v₁(t) + ½a(t)², why isnt Δx = x2-x1 = (v₁(t2) + ½a(t2)²) - (v₁(t1) + ½a(t1)²) = v₁(t₂ - t₁) + ½a(t₂² - t₁²) Why is it instead v₁(t₂ - t₁) + ½a(t₂ - t₁)²