A parallel plate capacitor of capacitance (air) is charged to , and the battery is then disconnected. A dielectric slab of and thickness equal to half the plate separation is slipped in between the plates (the other half remains air). The final stored energy is
Answer and explanation
Correct answer: C.
- Step 1: Prior to inserting the dielectric slab, the isolated capacitor holds charge .
- Step 2: Inserting a dielectric slab of thickness creates two capacitors in series: air of thickness () and dielectric of thickness (). The new capacitance is .
- Step 3: Because the battery was disconnected, charge is conserved. The final stored electrostatic energy is .
Answer
.
Why not the others: is the initial stored energy before inserting the dielectric slab; incorrectly assumes the voltage remains constant at after battery disconnection (); multiplies initial energy by dielectric constant .
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