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If qf is the bound charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then qb can be expressed as

1. qb = qf (1 - 1/√k)
2. qb = qf (1 - 1/k)
3. qb = qf (1 + 1/√k)
4. qb = qf (1 + 1/k)

Correct answer: qb = qf (1 - 1/k)

Solution

For a dielectric of constant k between capacitor plates, the bound (induced) charge is qb = qf(1 - 1/k), where qf is the free charge. The factor is always less than 1, so the answer is option 1.

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