AP Physics C E and M 9.2: Potential and Electric Field

Describe the relationship between electric potential and electric field, including how spatial changes in potential determine field direction and magnitude.

[Maximum number: 2]

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.

Derive an expression for the absolute value $|\Delta V|$ of the potential difference between

the outer and inner shells in terms of $R_{1}, R_{2}, \sigma_{1}$ , and physical constants, as appropriate.

Begin your derivation by writing a fundamental physics principle or an equation from

the reference information.

(ii) For substituting the expression for E from part A (i) into $\Delta V=-\int_{a}^{b} \vec{E} \cdot d \vec{r}$

Point A4

Scoring Notes:

- Vector notation is not required for this point to be earned.

- The sign of $\Delta V$ is not considered for this point to be earned.

For an attempt to solve the integral for $|\Delta V|$ that includes correct limits

Point A5

(e.g., $|\Delta V|=\frac{\sigma_{1} R_{1}}{\varepsilon_{0}} \int_{R_{1}}^{R_{2}} \frac{1}{r} d r$ )

Scoring Note: This point may be earned regardless of the order of the limits of

integration; for example, $\frac{\sigma_{1} R_{1}}{\varepsilon_{0}} \int_{R_{2}}^{R_{1}} \frac{1}{r} d r$ .

Example Response

$\begin{aligned} & |\Delta V|=\left|-\int_{a}^{b} \vec{E} \cdot d \vec{r}\right| \\ & |\Delta V|=\int_{R_{1}}^{R_{2}} \frac{\sigma_{1} R_{1}}{r \varepsilon_{0}} d r=\frac{\sigma_{1} R_{1}}{\varepsilon_{0}} \int_{R_{1}}^{R_{2}} \frac{1}{r} d r \\ & |\Delta V|=\left.\frac{\sigma_{1} R_{1}}{\varepsilon_{0}} \ln (r)\right|_{R_{1}} ^{R_{2}}=\frac{\sigma_{1} R_{1}}{\varepsilon_{0}} \ln \left(\frac{R_{2}}{R_{1}}\right) \end{aligned}$

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9.2.B—Describe the relationship between electric potential and electric field question 1