Answer:
a. 34.45 V
b. 89.80 pJ
c. 355.57 pJ
d. -265.67 pJ
Explanation:
(a) the potential difference between the plates
Since charge is conserved, charge before separation, Q = charge after separation, Q'
Q = C₁V₁ and C₁ = capacitance before separation = ε₀A/d₁ where A = area of plates = 6.70 cm² = 6.70 × 10⁻⁴ m²and d₁ = initial separation of plates = 2.50 mm = 2.50 × 10⁻³ m, V₁ = initial potential difference across plates = 8.70 V
Q' = C₂V₂ and C₂ = capacitance before separation = ε₀A/d₂ where A = area of plates = 6.70 cm² = 6.70 × 10⁻⁴ m² and d₁ = final separation of plates = 9.90 mm = 9.90 × 10⁻³ m. V₂ = final potential difference across plates = unknown
So, Q = Q'
C₁V₁ = C₂V₂
ε₀AV₁/d₁ = ε₀AV₂/d₂
So, V₁/d₁ = V₂/d₂
V₂ = V₁d₂/d₁
Substituting the values of the variables into the equation, we have
V₂ = 8.70 V × 9.90 mm/2.50 mm
V₂ = 8.70 V × 3.96
V₂ = 34.452 V
V₂ ≅ 34.45 V
(b) the initial stored energy
The energy stored in a capacitor U = 1/2CV².
So, the initial energy stored U₁ = 1/2C₁V₁² = ε₀AV₁²/2d₁
Substituting the values of the variables into the equation, we have
U₁ = ε₀AV₁²/2d₁
U₁ = 8.854 × 10⁻¹² F/m × 6.70 × 10⁻⁴ m² × (8.70V)²/(2 × 2.50 × 10⁻³ m)
U₁ = 4490.07 × 10⁻¹⁶ FmV²/5 × 10⁻³ m
U₁ = 898.01 × 10⁻¹³ J
U₁ = 89.801 × 10⁻¹² J
U₁ = 89.801 pJ
U₁ ≅ 89.80 pJ
(c) the final stored energy
The energy stored in a capacitor U = 1/2CV².
So, the initial energy stored U₂ = 1/2C₂V₂² = ε₀AV₂²/2d₂
Substituting the values of the variables into the equation, we have
U₂ = ε₀AV₂²/2d₂
U₂ = 8.854 × 10⁻¹² F/m × 6.70 × 10⁻⁴ m² × (34.45 V)²/(2 × 9.90 × 10⁻³ m)
U₂ = 70403.26 × 10⁻¹⁶ FmV²/19.8 × 10⁻³ m
U₂ = 3555.72 × 10⁻¹³ J
U₂ = 355.572 × 10⁻¹² J
U₂ = 355.572 pJ
U₂ ≅ 355.57 pJ
(d) the work required to separate the plates.
The work required to separate the plates W = -ΔU is the difference between the energy stored at the separation of 9.90 mm and 2.50 mm.
So, W = -(U₂ - U₁) = -(355.57 pJ - 89.80 pJ) = -265.67 pJ.
