P20B2014 – Three False Statements
Three False Statements
Three students are discussing the case of a point charge $+2Q$ surrounded by a spherical concentric metal shell with a net charge $+6Q$. These objects are fixed, and you may assume that $Q\gg e$.
They each have a theory about what happens when an electron is released from rest at point A located halfway between the center and the inside of the shell.
Terry says it won’t move since the electric field is zero because A is inside a conductor.
Susan says it will move radially outward since t is more attracted to the $+6Q$ charge on the inner surface of the shell.
Aaron says it will move radially inward since the net electric field at A points radially inward.
1. Briefly explain why each is wrong and use Gauss’s Law to help explain what really happens.
View answerHide answerTo determine what happens to the electron when it is released from rest at point A, we need to determine the electric field at point A.
Given the spherical symmetry of the problem, we will apply Gauss’s law to derive the electric field at point A. To do so we choose a Gaussian surface on which the magnitude of the electric field is constant and such that the electric field is perpendicular to the surface. Thus we choose a sphere centered on the point charge with a radius $r$ such that point A is a point of the surface of the sphere (as shown below).
The electric flux through the Gaussian surface is equal to
\oiint_{sphere}{\overrightarrow{E}\cdot d\overrightarrow{A}}=\oiint_{sphere}{E{\mathrm{cos} \left(0\right)\ }dA}=E\cdot 4\pi r^2The charge enclosed by the Gaussian surface is equal to the total charge of the inner point charge and we write
Q_{enc}=2QBy Gauss’s Law, we derive that
\begin{aligned}
\oiint_{sphere}{\overrightarrow{E}\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}\ \ \ \ &\Rightarrow \ \ \ \ \ E\cdot 4\pi r^2=\frac{2Q}{{\varepsilon }_0} \\
\\
&\Rightarrow \ \ \ \ \ E=\frac{2Q}{4\pi {\varepsilon }_0r^2}=\frac{Q}{2\pi {\varepsilon }_0r^2}
\end{aligned}The electric field at point $A$ is radially outward and has magnitude
E=\frac{Q}{2\pi {\varepsilon }_0r^2}Drawing the electric field at point $A$, we derive that the electric force acting on the electron at point $A$ is directed radially inward, toward the point charge $2Q$.
Terry is wrong because although point A is INSIDE a conducting shell, it is not INSIDE CONDUCTING MATERIAL and there is no reason for the electric field at point A to be zero (and it isn’t).
Susan is wrong because if the electric field is radially outward, then the electric force will pull the electron radially inward since the electron has a negative charge. Therefore, the electron will move radially inward.
Aaron is right about the direction in which the electron will move but gives a wrong reason. Indeed, the electric field is radially outward and not inward. It seems that he did not account for the sign of the charge of the electron!