P20B2019 – Long Wire Inside a Metal Tube
Long Wire Inside a Metal Tube
A long wire is surrounded by a long, thin, concentric, cylindrical metal tube of radius $R$. The wire has linear charge density $+\lambda $ and the tube has a net linear charge density $+2\lambda $.
A. Multiple Choice
An electron is released inside the tube at a radial distance $r_{elec}=R/2$ from the center wire. The direction of the force on the electron is
- a. radially inward
- b. no force on the electron
- c. radially outward
- d. parallel to the long wire
Answer: a
At $r_{elec}=R/2$ the electron only feels the electric field from the wire which is directed radially outward since the wire is positively charged. However, the electron being negative, the electric force is directed toward the wire and the electron moves radially inward.
B. In a separate experiment, a proton is released outside the tube at a radial distance $r_{prot}$ (measured from the center wire). The force on the proton has the same magnitude as the force on the electron described in Part A. Find $r_{prot}$, and the force on the proton (magnitude and direction). Justify your answers.
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Electric force acting on the electron:
Recalling that the electric field created by an infinite wire of charge with linear charge density $\lambda $ is equal to
E_w=\frac{\lambda }{2\pi {\varepsilon }_0r}we derive that the electric force exerted on the electron at location $r_{elec}=R/2$ has magnitude equal to
F_E=\left|-e\right|E_w=\frac{e\lambda }{2\pi {\varepsilon }_0R/2}=\frac{e\lambda }{\pi {\varepsilon }_0R}Electric field at location $r=r_{prot}$ :
To find the electric field at location $r=r_{prot}$ we consider a Gaussian cylinder of radius $r_{prot}$ and length $L$ centered on the wire as shown below.
The electric flux through the Gaussian cylinder is equal to
\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=\oiint_A{E{\mathrm{cos} \left(0\right)\ }dA}=E\cdot 2\pi rLThe charge enclosed by the Gaussian cylinder is equal to
Q_{enc}=\lambda L+2\lambda L=3\lambda LThus, by Gauss’s Law, the electric field at $r=r_{prot}$ has magnitude $E_p$ given by
\begin{aligned}
\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}\ \ \ \ \ &\Rightarrow \ \ \ \ \ E_p\cdot 2\pi r_{prot}L=\frac{3\lambda L}{{\varepsilon }_0} \\
\\ &\Rightarrow \ \ \ \ \ E_p=\frac{3\lambda }{2\pi {\varepsilon }_0r_{prot}}
\end{aligned}Electric force acting on the proton:
The force on the proton at location $r=r_{prot}$ is equal to
F_p=eE_p=\frac{3\lambda e}{2\pi {\varepsilon }_0r_{prot}}Setting the forces $F_e$ and $F_p$ equal to each other, we solve for $r_{prot}$ as follows
\frac{e\lambda }{\pi {\varepsilon }_0R}=\frac{3\lambda e}{2\pi {\varepsilon }_0r_{prot}}\ \ \ \ \Rightarrow \ \ \ \ \ r_{prot}=\frac{3R}{2}The distance $r_{prot}$ is therefore equal to
\boxed{r_{prot}=\frac{3R}{2}}C. What is the linear charge density on the outside of the long tube? Justify your answer.
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The linear charge density $\lambda $ of the wire pulls onto the inner surface of the cylinder a linear charge density ${\lambda }_{inner}=-\lambda $. Since the net linear charge density on the cylinder is ${\lambda }_{net}=2\lambda $, we conclude by conservation of charge that the outer surface of the cylinder carries a linear charge density equal to
\boxed{{\lambda }_{outer}={\lambda }_{net}-\left({\lambda }_{inner}\right)=2\lambda +\lambda =3\lambda}