1. Concept: Assuming the charged capacitor is isolated (disconnected from a battery), removing the dielectric removes the internal opposing polarization field. Thus, the total net field rebounds to its original vacuum strength.
2. Formula: $$ E_{vac} = E_{med} \times \epsilon_r $$
3. Solution: - While the dielectric was inserted, the electric field was suppressed (\( E_{med} = E_{vac} / \epsilon_r \)).
- Removing the dielectric means \( \epsilon_r \) effectively returns to 1.
- Because the trapped charge \( Q \) cannot escape, the lack of polarization mathematically forces the net electric field \( E \) to instantly increase back to its undamped vacuum state.
4. Why other options are incorrect: Option B is exactly what happens when you
insert the dielectric. Option D assumes the dielectric had no physical effect at all.
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