Tutorial 2 · Problem 6 — Force on Q₄ from 3 charges (square arrangement)
Given: Q₁=+25nC, Q₂=−25nC, Q₃=−25nC, Q₄=+25nC arranged in square (side=2, center on axes). Q₁ at (−2.5,−1), Q₂ at (−2.5,+1), Q₃ at (+2.5,−1), Q₄ at (+2.5,+1). Find F on Q₄.
R₁₄ = Q₄ − Q₁ = 5âx + 2ây, |R₁₄| = √(25+4) = √29 ... (adjusting for actual coordinates)
Using actual problem coordinates from tutorial:
R₁₄ = 2.5âx + 1ây, |R₁₄|² = 7.25
R₂₄ = 2.5âx − 1ây, |R₂₄|² = 7.25
R₃₄ = −2.5âx + 1ây, |R₃₄|² = 7.25
x = (−8.4 + 9.525)/2 = 0.5624 m from q₁ (take positive root only — must be between charges) x = −8.962 m is rejected (outside the range)
Tutorial 2 · Problem 7b — What happens if q₃ is slightly displaced?
Question: If q₃ is slightly moved right, what direction is the net force?
At equilibrium x=0.5624m: F₁₃ = F₂₃ (balanced).
If moved right slightly (x → 0.57m):
Distance to q₂ decreases → F₂₃ increases
Distance to q₁ increases → F₁₃ decreases
Net force → rightward (same direction as displacement)
Check numerically at x=0.57: F₂₃ > F₁₃ ✓
At x=0.56: F₁₃ > F₂₃ → net force leftward
This is an UNSTABLE equilibrium — any displacement grows!
Net force follows the displacement → unstable equilibrium Moved right → net force right. Moved left → net force left.
Tutorial 2 · Problem 7c — Does sign/magnitude of q₃ matter?
Question: If we change magnitude or sign of q₃, do we get a different position?
The equilibrium condition: q₁/(x²) = q₂/(1.2−x)²
Notice: q₃ cancels out from both sides!
The position x depends only on q₁ and q₂, NOT on q₃.
No — changing q₃ (sign or magnitude) does NOT change the equilibrium position. The position depends only on q₁ and q₂.
Lecture 4 · MCQ Example — Zero E field position
Given: 40C charge at x=4cm. Where to place −60C charge so E=0 at origin?
E at origin from +40C (at x=4cm): points in −x direction (toward charge: origin is left of charge, so field points left → but +Q field points AWAY → field at origin points in −x direction... wait: origin is to the LEFT of x=4cm. E from +Q points away from Q → points in −x direction at origin).
For E=0 at origin, the −60C must create equal magnitude E in +x direction.
Let −60C be at x = d (could be negative = left of origin).
E from −60C at origin points TOWARD it = in +x if d < 0.
Let d be negative: |d| = distance from origin to charge.
40/r₁² = 60/r₂² where r₁=0.04m, r₂=|d|
40/(0.04)² = 60/d² → d² = 60×(0.04)²/40 = 60×0.0016/40 = 0.0024
d = 0.049m = 4.9cm
Since charge must be at negative x: x = −4.9cm
Answer: (c) −4.9 cm (left of origin, on negative x-axis)
Lecture 4 · MCQ Example — E field at point P from two charges
Given: a=60cm, b=80cm, Q=−6nC, q=3nC. Find |E| at P.
Q at (0,0), q at (a,0)=(0.6,0), P at (0,b)=(0,0.8).
From Q=−6nC at origin: R = Pˉ−origin = 0.8ây, |R|=0.8m
E_Q = Q×R/(4πε₀|R|³) = (−6×10⁻⁹×0.8ây)/(4πε₀×0.512) → toward Q (−y direction since Q neg)
|E_Q| = 6×10⁻⁹/(4π×8.854×10⁻¹²×0.64) = 6×10⁻⁹/7.118×10⁻¹¹ = 84.3 N/C in −y
From q=3nC at (0.6,0): R = (−0.6âx+0.8ây), |R|=√(0.36+0.64)=1m
|E_q| = 3×10⁻⁹/(4πε₀×1) = 26.95 N/C
E_q = 26.95×(−0.6âx+0.8ây) = −16.17âx + 21.56ây