Math / Calculus

QuestionNote: Figure not drawn to scale.
A long insulating tube of length $L$ has inner radius $r_{a}$ and outer radius $r_{b}$ , where $L \gg r_{b}$ , as shown in the figure. The tube has a nonuniform charge distributed throughout its volume. The charge density $\rho$ as a function of the distance $r$ from the tube's central axis is $\rho(r)=\beta r^{3}$ for $r_{a}<r<r_{b}$ , where $\beta$ is a positive constant. lgnoring edge effects, what is the electric field $E$ in the range $r_{a}<r<r_{b}$ inside the tube? (A) 0 (B) $\frac{\beta}{4 \varepsilon_{a} r}\left(r^{4}-r_{a}^{4}\right)$ (C) $\frac{\beta}{5 \varepsilon_{0} r}\left(r^{5}-r_{a}^{5}\right)$ (D) $\frac{\beta}{5 \varepsilon_{0}{ }^{r}}\left(r_{b}^{5}-r^{5}\right)$ Question 6 of 24 Back Next

Studdy Solution

STEP 1

1. The tube is long, and edge effects can be ignored.
2. The charge density is given by $\rho(r) = \beta r^3$ for $r_a < r < r_b$ .
3. We are interested in finding the electric field $E$ for $r_a < r < r_b$ .

STEP 2

1. Use Gauss's Law to relate the electric field to the charge enclosed.
2. Calculate the charge enclosed within a Gaussian surface.
3. Solve for the electric field $E$ .

STEP 3

Use Gauss's Law, which states:
$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$
where $Q_{\text{enc}}$ is the charge enclosed by the Gaussian surface.

STEP 4

Consider a cylindrical Gaussian surface of radius $r$ and length $L$ , where $r_a < r < r_b$ . The charge enclosed, $Q_{\text{enc}}$ , is given by integrating the charge density over the volume:
$Q_{\text{enc}} = \int_{r_a}^{r} \rho(r') \cdot 2\pi r' L \, dr'$
Substitute $\rho(r') = \beta (r')^3$ :
$Q_{\text{enc}} = \int_{r_a}^{r} \beta (r')^3 \cdot 2\pi r' L \, dr'$
$= 2\pi \beta L \int_{r_a}^{r} (r')^4 \, dr'$

STEP 5

Evaluate the integral:
$\int_{r_a}^{r} (r')^4 \, dr' = \left[ \frac{(r')^5}{5} \right]_{r_a}^{r}$
$= \frac{r^5}{5} - \frac{r_a^5}{5}$
Thus, the charge enclosed is:
$Q_{\text{enc}} = 2\pi \beta L \left( \frac{r^5}{5} - \frac{r_a^5}{5} \right)$

STEP 6

Apply Gauss's Law to solve for $E$ :
$E \cdot 2\pi r L = \frac{2\pi \beta L}{\varepsilon_0} \left( \frac{r^5}{5} - \frac{r_a^5}{5} \right)$
$E = \frac{\beta}{5 \varepsilon_0 r} \left( r^5 - r_a^5 \right)$
The electric field $E$ in the range $r_a < r < r_b$ is:
$\boxed{\frac{\beta}{5 \varepsilon_0 r} \left( r^5 - r_a^5 \right)}$

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms