Chapter 22 Problem 9

Problem

${\vec {E}}$ muffler device

In a certain region of space, the electric field is constant in direction (say horizontal, in the ${\vec {x}}$ direction), but its magnitude decreases from $E=560{\textrm {N}}/{\textrm {C}}$ at $x=0$ to $E=410{\textrm {N}}/{\textrm {C}}$ at $x=25m$. Determine the charge within a cubical box of side $l=25{\textrm {m}}$, where the box is oriented so that four of its sides are parallel to the field lines

Solution

Out of 6 faces the cube has, only the left and the right face contributes, since the electric field, ${\vec {E}}$, is perpendicular to the surface normal, ${\hat {n}}$ (${\vec {E}}\cdot {\hat {n}}=0$ for these 4 surfaces)

The surface normal on the left hand side is towards left ($-{\hat {x}}$), and on the right hand side is right (${\hat {x}}$)

Total flux is

$\Phi _{E}=\left({\vec {E}}\cdot {\vec {A}}\right)_{\textrm {right}}+\left({\vec {E}}\cdot {\vec {A}}\right)_{\textrm {left}}+0+0+0+0$

The Gauss's law states the total flux through the surfaces of an enclosed volume is proportional to the charge it encapsulates

$\Phi _{E}=E_{\textrm {right}}l^{2}-E_{\textrm {left}}l^{2}={\frac {Q_{\textrm {encl}}}{\epsilon _{0}}}$

so the enclosed charge should be

$Q_{\textrm {encl}}=\left(E_{\textrm {right}}-E_{\textrm {left}}\right)l^{2}\epsilon _{0}=\left(410{\textrm {N}}/{\textrm {C}}-560{\textrm {N}}/{\textrm {C}}\right)\left(25{\textrm {m}}\right)^{2}\left(8.85\times 10^{-12}{\textrm {C}}^{2}/{\textrm {N}}{\textrm {m}}^{2}\right)=-8.3\times 10^{-7}{\textrm {C}}$