Question 23.7: A Cylindrically Symmetric Charge Distribution Find the elect...
A Cylindrically Symmetric Charge Distribution
Find the electric field a distance r from a line of positive charge of infinite length and constant charge per unit length λ (Fig. 23.16a, page 628).
Conceptualize The line of charge is infinitely long. Therefore, the field is the same at all points equidistant from the line, regardless of the vertical position of the point in Figure 23.16a. We expect the field to become weaker as we move farther away radially from the line of charge.
Categorize Because the charge is distributed uniformly along the line, the charge distribution has cylindrical symmetry and we can apply Gauss’s law to find the electric field.
Analyze The symmetry of the charge distribution requires that \overrightarrow{E} be perpendicular to the line charge and directed outward as shown in Figure 23.16b. To reflect the symmetry of the charge distribution, let’s choose a cylindrical gaussian surface of radius r and length \ell that is coaxial with the line charge. For the curved part of this surface, \overrightarrow{E} is constant in magnitude and perpendicular to the surface at each point, satisfying conditions (1) and (2). Furthermore, the flux through the ends of the gaussian cylinder is zero because \overrightarrow{E} is parallel to these surfaces. That is the first application we have seen of condition (3).
We must take the surface integral in Gauss’s law over the entire gaussian surface. Because \overrightarrow{E} \cdot d \overrightarrow{A} is zero for the flat ends of the cylinder, however, we restrict our attention to only the curved surface of the cylinder.
Apply Gauss’s law and conditions (1) and (2) for the curved surface, noting that the total charge inside our gaussian surface is \lambda \ell :
\Phi_E=\oint \overrightarrow{E} \cdot d \overrightarrow{A}=E \oint d A=E A=\frac{q_{\text {in }}}{\epsilon_0}=\frac{\lambda \ell}{\epsilon_0}Substitute the area A=2 \pi r \ell of the curved surface:
E(2 \pi r \ell)=\frac{\lambda \ell}{\epsilon_0}Solve for the magnitude of the electric field:
E=\frac{\lambda}{2 \pi \epsilon_0 r}=2 k_e \frac{\lambda}{r} (23.8)
Finalize This result shows that the electric field due to a cylindrically symmetric charge distribution varies as 1/r, whereas the field external to a spherically symmetric charge distribution varies as 1/r². Equation 23.8 can also be derived by direct integration over the charge distribution. (See Problem 8.)