---
title: "A parallel-plate capacitor with vacuum capacitance \\(C_0\\) is connected to a battery that maintains a constant potential difference \\(V_0\\). A dielectric slab with dielectric constant \\(\\kappa > 1\\) initially fills the region between the plates. The dielectric slab is then slowly removed from the capacitor while the capacitor remains connected to the battery. What is the change in electrostatic energy stored in the capacitor, \\(\\Delta U = U_{\\text{final}} – U_{\\text{initial}}\\)?"
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url: "https://nerd-notes.com/ubq/118197/"
date_modified: "2026-08-04T08:08:05+00:00"
---

# A parallel-plate capacitor with vacuum capacitance \(C_0\) is connected to a battery that maintains a constant potential difference \(V_0\). A dielectric slab with dielectric constant \(\kappa > 1\) initially fills the region between the plates. The dielectric slab is then slowly removed from the capacitor while the capacitor remains connected to the battery. What is the change in electrostatic energy stored in the capacitor, \(\Delta U = U_{\text{final}} – U_{\text{initial}}\)?

A parallel-plate capacitor with vacuum capacitance \(C_0\) is connected to a battery that maintains a constant potential difference \(V_0\). A dielectric slab with dielectric constant \(\kappa > 1\) initially fills the region between the plates. The dielectric slab is then slowly removed from the capacitor while the capacitor remains connected to the battery. What is the change in electrostatic energy stored in the capacitor, \(\Delta U = U_{\text{final}} - U_{\text{initial}}\)?

- **A.** \(-\dfrac{1}{2}(\kappa - 1) C_0 V_0^2\)
- **B.** \(+\dfrac{1}{2}(\kappa - 1) C_0 V_0^2\)
- **C.** \(-\dfrac{1}{2}\left(\dfrac{\kappa - 1}{\kappa}\right) C_0 V_0^2\)
- **D.** \(-(\kappa - 1) C_0 V_0^2\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118197/*