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A parallel plate capacitor is connected across a battery and the energy stored in the capacitor is $E$ - AQA - A-Level Physics - Question 20 - 2022 - Paper 2
Question 20
A parallel plate capacitor is connected across a battery and the energy stored in the capacitor is $E$. Without disconnecting the battery, the separation of the plat... show full transcript
Worked Solution & Example Answer:A parallel plate capacitor is connected across a battery and the energy stored in the capacitor is $E$ - AQA - A-Level Physics - Question 20 - 2022 - Paper 2
Step 1
What is the initial energy stored in the capacitor?
Answer
The initial energy stored in a capacitor can be expressed using the formula: U = rac{1}{2} C V^2 where is the energy, is the capacitance, and is the voltage across the capacitor. Given that the stored energy is , we have: E = rac{1}{2} C V^2
Step 2
What happens when the plate separation is halved?
Answer
When the separation of the plates is halved, the capacitance changes. The capacitance of a parallel plate capacitor is given by: C = rac{ ext{ε} A}{d} where is the permittivity of the material between the plates, is the area of the plates, and is the separation between them. By halving , the new capacitance becomes: C' = rac{ ext{ε} A}{d/2} = 2 imes rac{ ext{ε} A}{d} = 2C
Step 3
What is the new energy stored in the capacitor?
Answer
Since the capacitor remains connected to the battery, the voltage across the capacitor does not change. Thus, we can calculate the new energy stored using the new capacitance: U' = rac{1}{2} C' V^2 = rac{1}{2} (2C) V^2 = 2 imes rac{1}{2} C V^2 = 2E Therefore, the energy now stored in the capacitor is .