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QPages 164-165: Q2: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 164-165: Q2 Solution: A dielectric slab is inserted between the plates of an isolated …
⚠️ Common Exam Traps & Conceptual Pitfalls
In an isolated capacitor the charge $Q$ is conserved; inserting a dielectric increases the capacitance $C$, so the voltage $V=Q/C$, the field $E=V/d$, and the stored energy $U= frac12 CV^2$ all decrease. Students often mistakenly assume $V$ or $E$ stay the same, which would only be true if the capacitor were connected to a voltage source rather than isolated.
A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
Multiple Choice Options
- A The electric field in the capacitor
- B The charge on the capacitor
- C The potential difference between the plates
- D The stored energy in the capacitor
Step-by-Step Verified Solution
**Answer:** (b) The charge on the capacitor.
**Step-by-Step Explanation:**
For an isolated capacitor, Q is constant. V, E, and U change.
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