P20B2017 – Infinite Wire of Unknown Charge

Answer: c

The linear charge density outside of the cylinder is $+4\lambda $ and therefore we conclude that the arrangement $\mathrm{{}$wire + cylinder$\mathrm{}}$ behaves like a single line of charge with linear charge density $+4\lambda $ when viewed from the outside. Therefore, we conclude that the electric field at a distance $r>1.2R$ will be radially outward.

The proton placed at $r=1.5R$ will therefore experience a force that is directed radially outward.

Answer: a

If the cylinder’s net linear charge density is $-8\lambda $ and the linear charge density on its outer surface is $+4\lambda $, then we conclude, by conservation of charge, that the linear charge density ${\lambda }_{inner}\ $ on its inner surface of the cylinder is given by

\begin{aligned}
{\lambda }_{inner}+{\lambda }_{outer}={\lambda }_{tot}\ \ \ \ &\Rightarrow \ \ \ \ \ {\lambda }_{inner}={\lambda }_{tot}-{\lambda }_{outer} \\
\\
&\Rightarrow \ \ \ \ \ \boxed{{\lambda }_{inner}=-8\lambda -4\lambda =-12\lambda}
\end{aligned}

Thus, the linear charge density of the wire is $+12\lambda $ and the wire is therefore positively charged.

The electric field created at $r=0.5R$ is the electric field created by the central wire which is directed radially outward and has magnitude

E=\frac{12\lambda }{2\pi {\varepsilon }_0r}=\frac{12\lambda }{\pi {\varepsilon }_0R}

The electric force acting on an electron placed at $r=0.5R$ is therefore directed radially inward and has magnitude

\boxed{F_E=\left|-e\right|E=\frac{12e\lambda }{\pi {\varepsilon }_0R}}

The electric field lines for this charge configuration are drawn below