Answer:
a. 4.19 J b. 16.76 J
Explanation:
Here is the complete question
A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plate is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?
Solution
The energy stored W in a capacitor of capacitance c and charge Q stored = W = Q²/2C.
Let C₀ be the initial capacitance at d₀ = 2.30 mm separation and C₁ be the initial capacitance at d₁ = 1.15 mm separation.
The initial charge stored in the capacitor equals Q₀ = √2W₀C₀
The energy stored at a separation of 1.15 mm = W₁ = Q₀²/2C₁. Since the charge is constant. Since C ∝ 1/d, (since A is constant) C₀/C₁ = d₁/d₀. So, C₁ = (d₀/d₁)C₀ = (2.30/1.15)C₀ = 2C₀
Therefore W₁ = Q₀²/2C₁ = (√2W₀C₀)²/2(2C₀) = 2W₀C₀/4C₀ = W₀/2 = 8.38 J/2 = 4.19 J
b. We know energy stored W = (1/2)CV² and V = √(2W/C)
The initial voltage V₀ = √(2W₀/C₀).
The new energy stored W₂ = (1/2)C₁V₀² (since the voltage is constant)
W₂ = (1/2)C₁V₀² =(1/2)(2C₀)(√(2W₀/C₀))² = C₀(2W₀/C₀) = 2W₀ = 2 × 8.38 J = 16.76 J
