---
title: "An isolated solid conductor in electrostatic equilibrium carries a uniform positive surface charge density \\(\\sigma\\). The electric field just outside the surface of the conductor has magnitude \\(E = \\dfrac{\\sigma}{\\varepsilon_0}\\) and points perpendicular to the surface. Which of the following correctly states the magnitude of the electrostatic force \\(dF\\) exerted on a small patch of area \\(dA\\) by all the other charges on the conductor, and provides the correct physical justification?"
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url: "https://nerd-notes.com/ubq/121141/"
date_modified: "2026-08-23T04:58:17+00:00"
---

# An isolated solid conductor in electrostatic equilibrium carries a uniform positive surface charge density \(\sigma\). The electric field just outside the surface of the conductor has magnitude \(E = \dfrac{\sigma}{\varepsilon_0}\) and points perpendicular to the surface. Which of the following correctly states the magnitude of the electrostatic force \(dF\) exerted on a small patch of area \(dA\) by all the other charges on the conductor, and provides the correct physical justification?

An isolated solid conductor in electrostatic equilibrium carries a uniform positive surface charge density \(\sigma\). The electric field just outside the surface of the conductor has magnitude \(E = \dfrac{\sigma}{\varepsilon_0}\) and points perpendicular to the surface. Which of the following correctly states the magnitude of the electrostatic force \(dF\) exerted on a small patch of area \(dA\) by all the other charges on the conductor, and provides the correct physical justification?

![A portion of a curved conducting surface shown in cross section with a thick solid boundary line. The region to the left of the boundary represents the interior of the conductor and contains the centered text label \(E = 0\). The region to the right of the boundary represents the space outside the conductor. A small segment of the boundary is marked by two short perpendicular tick marks and labeled \(dA\). An arrow labeled \(E = \dfrac{\sigma}{\varepsilon_0}\) starts at the outer surface and points perpendicularly outward to the right. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787461096-YQLf3j.jpg)

- **A.** \(dF = \dfrac{\sigma^2}{\varepsilon_0}\,dA\), because the charge \(dq = \sigma\,dA\) on the patch is immersed directly in the total electric field of magnitude \(E = \dfrac{\sigma}{\varepsilon_0}\) measured just outside the conductor.
- **B.** \(dF = \dfrac{\sigma^2}{2\varepsilon_0}\,dA\), because the patch does not exert a net force on itself, and the electric field produced exclusively by the remaining charges on the conductor has magnitude \(\dfrac{\sigma}{2\varepsilon_0}\) at the patch.
- **C.** \(dF = \dfrac{\sigma^2}{4\varepsilon_0}\,dA\), because the field produced by the remaining charges must split equally into internal and external halves, leaving only one-fourth of the surface flux to push the patch outward.
- **D.** \(dF = 0\), because the conductor is in electrostatic equilibrium, which requires the net electrostatic force acting on every local surface charge element to be zero.

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