---
title: "A parallel-plate capacitor with plate area \\(A\\) and plate separation \\(d\\) is charged so that the plates carry free charges \\(+Q_0\\) and \\(-Q_0\\), and is then disconnected from the voltage source. A slab of linear dielectric material with dielectric constant \\(\\kappa > 1\\) is inserted, completely filling the space between the plates. Which of the following statements and justifications correctly describes the magnitude of the net electric field \\(E\\) inside the capacitor after the dielectric is inserted?"
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url: "https://nerd-notes.com/ubq/121114/"
date_modified: "2026-08-23T04:58:08+00:00"
---

# A parallel-plate capacitor with plate area \(A\) and plate separation \(d\) is charged so that the plates carry free charges \(+Q_0\) and \(-Q_0\), and is then disconnected from the voltage source. A slab of linear dielectric material with dielectric constant \(\kappa > 1\) is inserted, completely filling the space between the plates. Which of the following statements and justifications correctly describes the magnitude of the net electric field \(E\) inside the capacitor after the dielectric is inserted?

A parallel-plate capacitor with plate area \(A\) and plate separation \(d\) is charged so that the plates carry free charges \(+Q_0\) and \(-Q_0\), and is then disconnected from the voltage source. A slab of linear dielectric material with dielectric constant \(\kappa > 1\) is inserted, completely filling the space between the plates. Which of the following statements and justifications correctly describes the magnitude of the net electric field \(E\) inside the capacitor after the dielectric is inserted?

![Two horizontal parallel conducting plates separated vertically by distance d. The top plate is labeled +Q_0 and the bottom plate is labeled -Q_0. A solid shaded rectangular slab labeled \kappa completely fills the space between the two plates. No wires or circuits are attached to the plates. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787461087-kyQ7vn.jpg)

- **A.** The electric field increases because the dielectric constant \(\kappa > 1\) increases the capacitance, drawing additional free charge onto the plates from within the dielectric.
- **B.** The electric field stays the same because the capacitor is isolated, so the free surface charge density on the conducting plates is unchanged and solely determines the total internal field.
- **C.** The electric field decreases because free electrons conduct across the dielectric to partially neutralize the free charge residing on the conducting plates.
- **D.** The electric field decreases because bound polarization charges on the dielectric surfaces create an induced electric field that opposes the field produced by the constant free charges on the plates.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121114/*