Question

The electric field $\overrightarrow{E_1}$ at one face of a parallelopiped is uniform over the entire face and is directed out of the face. At the opposite face, the electric field ${\overrightarrow{E}}_2$ is also uniform over the entire face and is directed into that face (as shown in Fig.). The two faces in question are inclined at ${30}^{\circ }$ from the horizontal, ${\overrightarrow{E}}_1$ and ${\overrightarrow{E}}_2$ (both horizontal) have magnitudes of $2.50\times {10}^{4}N/C$ and $7.00\times {10}^{4}N/C$, respectively. Assuming that no other electric field lines cross the surfaces of the parallelopiped, the net contained within is

• A. $-67.5{\epsilon }_{0}C$
• B. $37.5{\epsilon }_{0}C$
• C. $105{\epsilon }_{0}C$
• D. $-105{\epsilon }_{0}C$

Answer 1

Answer: (A)

$-67.5{\epsilon }_{0}C$

To find the charge enclosed, we need the flux through the parallelpiped:
${\Phi }_1=A{E}_{1}cos{60}^{\circ }$
$=\left(0.0500m\right)\left(0.0600m\right)(2.50\times {10}^{4}N{C}^{-1})cos{60}^{\circ }$
$=37.5N{m}^2{C}^{-1}$
${\Phi }_2=A{E}_{2}cos{120}^{\circ }$
$=\left(0.0500m\right)\left(0.0600m\right)(7.00\times {10}^{4}N{C}^{-1})cos{120}^{\circ }$
$=-105N{m}^2{C}^{-1}$
So, the total flux is
$\Phi ={\Phi }_1+{\Phi }_2=(37.5-105)N{m}^2{C}^{-1}=-67.5N{m}^2{C}^{-1}$
$q=\Phi {\epsilon }_0=(-67.5N{m}^2{C}^{-1}){\epsilon }_0=-5.97\times {10}^{-10}C$
There must be net charge (negative) in the parallelepiped since there is a
net flux flowing into the surface. Also, there must be an external
field, otherwise all lines would point toward the slab.