Question

A long, nonconducting cylinder (radius = 12 cm) has a charge of uniform density (5.0 nC/m3)
distributed throughout its volume. Determine the magnitude of the electric field 15 cm from
the axis of the cylinder.
a) 20 N/C
b) 27 N/C
c) 16 N/C
d) 12 N/C
e) 54 N/C

Answer 1

To determine the electric field at a distance from a long, nonconducting cylinder with uniform charge density, we can use Gauss's law.

**Given:**
- Charge density, \( \rho = 5.0 \, \text{nC/m}^3 = 5.0 \times 10^{-9} \, \text{C/m}^3 \)
- Radius of the cylinder, \( R = 12 \, \text{cm} = 0.12 \, \text{m} \)
- Distance from the axis at which we want to calculate the electric field, \( r = 15 \, \text{cm} = 0.15 \, \text{m} \)

Since \( r > R \), we can treat the cylinder as if it were a line of charge when calculating the electric field outside of the cylinder. The electric field due to a uniformly charged infinite cylinder outside its surface can be given by the formula:

\[
E = \frac{1}{2 \pi \epsilon_0} \cdot \frac{Q_{\text{inside}}}{r^2}
\]

**Step 1: Find total charge within the cylinder's cross-section.**

The charge contained in the cylinder's volume can be found as:
\[
Q = \rho V
\]

where
- \( V \) is the volume of the cylinder up to radius \( R \) and length \( L \):

\[
V = \pi R^2 L
\]

Assuming a length \( L \) of the cylinder, we have:

\[
Q = \rho \cdot (\pi R^2 L) = (5.0 \times 10^{-9} \, \text{C/m}^3) \cdot \pi (0.12 \, \text{m})^2 L
\]

Calculating \( \pi (0.12 \, \text{m})^2 \):

\[
\pi (0.12)^2 \approx \pi \cdot 0.0144 \approx 0.045238934 \, \text{m}^2
\]

Now calculating \( Q \):

\[
Q = (5.0 \times 10^{-9}) \cdot (0.045238934) \cdot L \approx 2.268 \times 10^{-10} L \, \text{C}
\]

**Step 2: Use Gauss's Law to find the electric field at \( r = 15 \, \text{cm} \).**

Choosing a Gaussian surface of a cylinder of radius \( r = 0.15 \, \text{m} \) and length \( L \), utilizing Gauss's law:

\[
\Phi_E = E(2 \pi r L) = \frac{Q_{\text{enc}}}{\epsilon_0}
\]

where \( Q_{\text{enc}} = Q \) from the earlier calculation.

Since the Gauss's law gives:

\[
E(2 \pi (0.15) L) = \frac{2.268 \times 10^{-10} L}{\epsilon_0}
\]

Simplifying for \( E \):

\[
E = \frac{2.268 \times 10^{-10}}{(2 \pi (0.15) \epsilon_0)}
\]

Using \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \)

Calculating \( E \):

\[
E = \frac{2.268 \times 10^{-10}}{(2 \pi (0.15) (8.85 \times 10^{-12}))}
\]

Calculating the denominator:

\[
2 \pi (0.15) (8.85 \times 10^{-12}) \approx 8.325659 \times 10^{-12}
\]

Thus,

\[
E \approx \frac{2.268 \times 10^{-10}}{8.325659 \times 10^{-12}} \approx 27.24 \, \text{N/C}
\]

**Final Result:**

The magnitude of the electric field 15 cm from the axis of the cylinder is approximately **27 N/C**.

So the answer is:
**b) 27 N/C**.