AP Physics C E and M 8.6 Gauss Law Overview

Apply Gauss’s law to connect electric flux through a closed surface with enclosed charge, using symmetry to determine electric fields.

[Maximum number: 3]

An isolated, air-filled, charged capacitor consists of two conducting, coaxial, cylindrical shells

that each have length L. The inner shell has radius $R_{1}$ and the outer shell has radius $R_{2}$ , as

shown in Figure 1, where $R_{1}<R_{2} \ll L$ . The surface charge densities (amounts of charge per unit

area) of the inner and outer shells are $+\sigma_{1}$ and $-\sigma_{2}$ , respectively. The absolute values of the

total charges on the shells are equal.

Figure 1

Note: Figures not drawn to scale.

A.

Using Gauss's law, derive an expression for the magnitude E of the electric field as

a function of the radial distance r from the center of the capacitor for the region

$R_{1}<r<R_{2}$ . Express your answer in terms of $R_{1}, \sigma_{1}, r$ , and physical constants, as

appropriate.

A (i)

For a multistep derivation that includes the equation $\oint \vec{E} \cdot d \vec{A}=\frac{q_{\text {enc }}}{\varepsilon_{0}}$

Point A1

Scoring Note: Vector notation is not required for this point to be earned.

For a correct substitution of the area of an appropriate Gaussian surface with nonzero

Point A2

flux for the region $R_{1}<r<R_{2}$ (e.g., $2 \pi r \ell$ )

For a correct expression for the enclosed charge (e.g., $\sigma_{1}\left(2 \pi R_{1} \ell\right)$ )

Point A3

Example Response

$\begin{aligned} & \oint \vec{E} \cdot d \vec{A}=\frac{q_{\text {enc }}}{\varepsilon_{0}} \\ & E(2 \pi r \ell)=\frac{\sigma_{1}\left(2 \pi R_{1} \ell\right)}{\varepsilon_{0}} \\ & E=\frac{\sigma_{1} R_{1}}{\varepsilon_{0} r} \end{aligned}$

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Covered syllabus points

8.6 Gauss’s Law question 1