University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

96

Pages:

6
Author:

Jabin F.

Potential at point P [V = \frac{1}{4\pi\epsilon_{o}}\left[\frac{q}{r_{+}}+\frac{q^{\prime}}{r_{-}}\right]] Law of Cosines: r_{+}=\sqrt{a^2+a^2-2a^2\cos\theta} r_{-}=\sqrt{a^2+b^2-2ab\cos\theta} Substitution: b = R^2/a Potential Calculation: [V = \frac{1}{4\pi\epsilon_{o}}\left[\frac{q}{\sqrt{a^2+a^2-2a^2\cos\theta}}+\frac{q^{\prime}}{\sqrt{a^2+(R^2/a)^2-2a(R^2/a)\cos\theta}}\right]] If charge at (2R, 0, 0): q' = -q/2 V = \frac{1}{4\pi\epsilon_{o}}\left[\frac{q}{R}-\frac{q/2}{\sqrt{R^2+R^2-2R^2\cos\theta}}\right] V = \frac{q}{4\pi\epsilon_{o}}\left[\frac{1}{R}-\frac{1}{2R}\right] = \frac{q}{8\pi\epsilon_{o}R} Ques-48 mg = qE/(4πε₀(2d)²) Ques-49 Diagram: Two charges, +q and -q, separated by a distance 2d. A point P is located at a distance d above the midpoint of the charges. Enet = Ec + Ed F = (+q)Enet Enet = 2/(4πε₀(2d)²) * (-x̂) + 2q/(4πε₀(2d)²) * (-ŷ) Enet = (q/(4πε₀(4d²))) * (-x̂ - ŷ + ẑ) Enet = (q/(4πε₀(4d²))) * √(2^2 + 1^2) Enet = q/(4πε₀(4d²)) * √5 τ = p × Enet = (qd) × (q/(4πε₀(4d²))) * √5 τ = q²/(4πε₀(4d)) * √5 (towards the corner) Ques. 58 |FA| = 2²/(32π²ε₀d²) |FB| = 16/(16π²ε₀d²) |FA|/|FB| = 2/16 = 1/8 = 0.125 Ques. 50 |FA| = 2²/(16π²ε₀d²) |FB| = (2q)²/(48π²ε₀d²) |FA|/|FB| = 9/(4d²) = 9/4 d'/d = √(9/4) = 3/2