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QPages 164-165: Q5: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 164-165: Q5 Solution: The separation between the plates of a charged parallel-plate…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the capacitor as if it remains connected to a voltage source, keeping $V$ fixed, when in fact an isolated capacitor has constant charge $Q$. Remember that for an isolated capacitor $Q$ is unchanged, so increasing the plate separation reduces $C$, which raises $V$, increases the stored energy $U=\frac{Q^2}{2C}$, and alters the energy density $u=\frac{1}{2}\varepsilon_0E^2$.

The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?

Multiple Choice Options

  • A Charge on the capacitor
  • B Potential difference across the capacitor
  • C Energy of the capacitor
  • D Energy density between the plates

Step-by-Step Verified Solution

**Answer:** (b), (c), (d). **Step-by-Step Explanation:** For an isolated capacitor, Q is constant. If separation increases, C decreases, V increases, U increases, and energy density changes.

All Questions in this Chapter

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