A parallel-plate capacitor with air between its plates has capacitance and is charged to potential . It is then disconnected from the battery, and a dielectric slab of is inserted filling the space. What are the new voltage and stored energy?
Answer and explanation
Correct answer: A. ,
- Step 1: Since disconnected, charge is conserved: .
- Step 2: New capacitance: .
- Step 3: New potential: .
- Step 4: New energy: .
Answer
V' = 20 V, U' = 0.010 J.
Why not the others: '' assumes battery remains connected; initial energy was (energy decreased by factor of ).
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