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.
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