Answer:
α = 141.5° (counterclockwise)
Explanation:
If
q₁ = +q
q₂ = -q
q₃ < 0
b = 2*a
We apply Coulomb's Law as follows
F₁₃ = K*q₁*q₃ / d₁₃² = + K*q*q₃ / (2*a)² = + K*q*q₃ / (4*a²)
F₂₃ = K*q₂*q₃ / d₂₃² = - K*q*q₃ / (5*a²)
(d₂₃² = a² + (2a)² = 5*a²)
Then
∅ = tan⁻¹(2a/a) = tan⁻¹(2) = 63.435°
we apply
F₃x = - F₂₃*Cos ∅ = - (K*q*q₃ / (5*a²))* Cos 63.435°
⇒ F₃x = - 0.0894*K*q*q₃ / a²
F₃y = - F₂₃*Sin ∅ + F₁₃
⇒ F₃y = - (K*q*q₃ / (5*a²))* Sin 63.435° + (K*q*q₃ / (4*a²))
⇒ F₃y = 0.0711*K*q*q₃ / a²
Now, we use the formula
α = tan⁻¹(F₃y / F₃x)
⇒ α = tan⁻¹((0.0711*K*q*q₃ / a²) / (- 0.0894*K*q*q₃ / a²)) = - 38.5°
The real angle is
α = 180° - 38.5° = 141.5° (counterclockwise)
Answer:
0.03605 V/m is the electric field in the gold wire.
Explanation:
Resistivity of the gold = 
Length of the gold wire = L = 14 cm = 0.14 m ( 1 cm = 0.01 m)
Diameter of the wire = d = 0.9 mm
Radius of the wire = r = 0.5 d = 0.5 × 0.9 mm = 0.45 mm = 
( 1mm = 0.001 m)
Area of the cross-section = 
Resistance of the wire = R
Current in the gold wire = 940 mA = 0.940 A ( 1 mA = 0.001 A)

( Ohm's law)

We know, Electric field is given by :




0.03605 V/m is the electric field in the gold wire.
Answer:
The answer is continuity ( D )
Explanation:
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