An imaginary cubical surface has one corner located at the origin of a cartesian frame with its sides parallel to the x-y, x-z, and y-z planes.
An electric field given by
E = (125 N/C·m)y
^
j
passes through the faces of the cube. The cube side has a length of
8.00 cm.
eo=8.85×10-12 C2/N·m2
(a) Calculate the total electric flux passing through the cube?
The only surface through which electric flux passes is the y=a plane.
fy=a
=
(125 N/C·m)a
^
j
· (a2)
^
j
=
(125 N/C·m)(0.0800 m)3
=
+6.40×10-2 N·m2/C
=
FE
(b) Calculate the net charge inside the cube (magnitude and sign).
qin
=
eoFE
=
(+6.40×10-2 N·m2/C) (8.85×10-12 C2/N·m2)
=
+5.66×10-13 C
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version 3.85. On 03 Feb 2016, 11:34.