what will be the work done if a Unit chargeis moved from infinity to a distance 'R' from infinitely large charged sheet? CUT TO
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Answer:
Explanation:
In the x direction the force will be
½(-w₀)L/2 = -¼w₀L
acting ⅔(L/2) = L/3 below the x axis.
In the y direction the force will be
½(-w₀)L + ½w₀L/2 = -¼w₀L
the magnitude of the resultant will be
F = w₀L √((-¼)² + (-¼)²) = w₀L√⅛
in the direction
θ = arctan(-¼w₀L / -¼w₀L) = 225°
to find the distance, we balance moments
(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]
(√⅛)[d] = ½ [⅔L] + ¼ [⅔L/2] - ¼ [L - ⅓L/2]
(√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]
(√⅛)[d] = ⅓L + ⅟₁₂L - ¼L + ⅟₂₄L
(√⅛)[d] = 5L/24
d = 5L/24 / (√⅛)
d = 5√⅛L/3
