A small, rigid object carries positive and negative 2.00 nC charges. It is orient [Free] B116

A small, rigid object carries positive and negative 2.00 nC charges. It is oriented so that the positive charge has coordinates (-1.20mm,1.60mm)1.50mm,-1.30mm Cmhat(i)+,Cmhat(j) (b) The object is placed in an electric field vec(E)=(7.80×103(hat(i))-4.90×103(hat(j))) NC.Find the torque acting on the object. N*m—Select— v (c) Find the potential energy of the object-field system when the object is in this orientation. J (d) Assuming the orientation of the object can change, find the difference between the maximum and the minimum potential energies of the system. J

A small, rigid object carries positive and negative 2.00 nC charges. It is orient [Free] B116

Answer

Electric Dipole Moment, Torque & Potential Energy – Step-by-Step Guide

Electric Dipole Moment, Torque & Potential Energy

Step 1: Electric Dipole Moment (⃗p)

Given:

  • Charge: q = 2.00 × 10⁻⁹ C
  • Positive charge location: (-1.20 × 10⁻³, 1.60 × 10⁻³) m
  • Negative charge location: (1.50 × 10⁻³, -1.30 × 10⁻³) m

Formula:

⃗p = q × ⃗d

Where ⃗d is the vector from negative to positive charge.

Displacement vector:

⃗d = ⃗r₊ – ⃗r₋ = (-2.70 × 10⁻³, 2.90 × 10⁻³) m

Dipole moment:

⃗p = (2.00 × 10⁻⁹) × (-2.70 × 10⁻³, 2.90 × 10⁻³)
⃗p = (-5.40 × 10⁻¹², 5.80 × 10⁻¹²) C·m

Final Result: ⃗p = -5.40 × 10⁻¹² î + 5.80 × 10⁻¹² ĵ C·m

Step 2: Torque on the Dipole (τ⃗)

Given Electric Field: ⃗E = (7.80 × 10³, -4.90 × 10³) N/C

Formula:

τ⃗ = ⃗p × ⃗E

Cross product in 2D:

τ = (pₓ × Eᵧ – pᵧ × Eₓ)
= (-5.40 × 10⁻¹²)(-4.90 × 10³) – (5.80 × 10⁻¹²)(7.80 × 10³)
= 2.646 × 10⁻⁸ – 4.524 × 10⁻⁸
= -1.878 × 10⁻⁸ N·m

Final Result: τ = -1.878 × 10⁻⁸ N·m

Step 3: Potential Energy of Dipole in Electric Field

Formula:

U = -⃗p · ⃗E

Dot product:

U = -[(-5.40 × 10⁻¹²)(7.80 × 10³) + (5.80 × 10⁻¹²)(-4.90 × 10³)]
= -[-4.212 × 10⁻⁸ – 2.842 × 10⁻⁸]
= 7.054 × 10⁻⁸ J

Final Result: U = 7.054 × 10⁻⁸ J

Step 4: Maximum and Minimum Potential Energy

Formulas:

  • Umax = +|⃗p||⃗E|
  • Umin = -|⃗p||⃗E|
  • ΔU = Umax – Umin

Magnitude of Dipole Moment:

|⃗p| = √((-5.40)² + (5.80)²) × 10⁻¹²
= √62.80 × 10⁻¹² ≈ 7.93 × 10⁻¹² C·m

Magnitude of Electric Field:

|⃗E| = √((7.80)² + (-4.90)²) × 10³
= √84.85 × 10³ ≈ 9.21 × 10³ N/C

Maximum Potential Energy:

Umax = (7.93 × 10⁻¹²) × (9.21 × 10³) = 7.423 × 10⁻⁸ J

Energy Range:

ΔU = 2 × Umax = 2 × 7.423 × 10⁻⁸ = 1.485 × 10⁻⁷ J

Final Result:

  • Umax = 7.423 × 10⁻⁸ J
  • ΔU = 1.485 × 10⁻⁷ J

Final Summary

  • Dipole Moment: ⃗p = -5.4010 × 10⁻¹² î + 5.8010 × 10⁻¹² ĵ C·m
  • Torque: τ = -1.878 × 10⁻⁸ N·m
  • Potential Energy: U = 7.05410 × 10⁻⁸ J
  • Energy Range: ΔU = 1.485 × 10⁻⁷ J

Similar Posts

Leave a Reply

Your email address will not be published. Required fields are marked *