how much work is required to completely separate two charges (each -1.4 microcoulombs) and leave them at rest if they were initially 8 mm apart?
To calculate the work required to completely separate two charges, we can use the formula:
W = ΔU = q₁q₂ / (4πε₀r)
Where:
W is the work done
ΔU is the change in potential energy
q₁ and q₂ are the charges
ε₀ is the permittivity of free space (8.854 × 10⁻¹² C²/(N·m²))
r is the initial separation between the charges
Given:
q₁ = q₂ = -1.4 μC (converting to Coulombs: -1.4 × 10⁻⁶ C)
r = 8 mm (converting to meters: 8 × 10⁻³ m)
Now we can calculate the work:
W = (-1.4 × 10⁻⁶ C)(-1.4 × 10⁻⁶ C) / (4π(8.854 × 10⁻¹² C²/(N·m²))(8 × 10⁻³ m))
W = (1.96 × 10⁻¹² C²) / (4π(8.854 × 10⁻¹² C²/(N·m²))(8 × 10⁻³ m))
W = (1.96 × 10⁻¹²) / (4π(8.854 × 10⁻¹²))(8 × 10⁻³)
W ≈ 7.891 × 10⁻¹⁰ J
Therefore, the work required to completely separate the two charges and leave them at rest is approximately 7.891 × 10⁻¹⁰ Joules.
To calculate the work required to separate two charges, we can use the formula for electrical potential energy:
\(W = PE = \frac{{k \cdot q_1 \cdot q_2}}{{d}}\)
Where:
- \(W\) is the work done (in Joules)
- \(PE\) is the potential energy
- \(k\) is the electrostatic constant (\(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\))
- \(q_1\) and \(q_2\) are the charges (in Coulombs)
- \(d\) is the distance between the charges (in meters)
Given that each charge is -1.4 microcoulombs (\(-1.4 \times 10^{-6}\, \text{C}\)) and the initial separation is 8 mm (or 0.008 m), we can substitute these values into the formula:
\(W = \frac{{(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) \cdot (-1.4 \times 10^{-6} \, \text{C})^2}}{{0.008 \, \text{m}}}\)
Now we can solve this equation to find the work required.