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A parallel plate capacitor has capacitance C with air between the plates. The space between the plates is then completely filled with a dielectric slab of dielectric constant k, and the capacitor is connected to a cell of emf E. Subsequently, the dielectric slab is slowly pulled out. Which of the following statements are correct?

1. A charge of CE(k-1) flows through the cell during the removal of the slab.
2. The cell absorbs energy equal to E²*C*(k-1) during the removal of the slab.
3. The energy stored in the capacitor decreases by E²*C*(k-1)/2 when the slab is removed.
4. The external agent must do (1/2)*E²*C*(k-1) work to remove the slab.

Correct answer: A charge of CE(k-1) flows through the cell during the removal of the slab.

Solution

As the slab is removed, charge decreases from kCE to CE, so CE(k-1) flows back through the cell (cell absorbs this charge, gaining energy CE²(k-1)). The stored energy decreases by (k-1)CE²/2. By energy conservation, work by external agent = change in stored energy - energy returned to cell = -(k-1)CE²/2... let's verify carefully.

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