The figure at right shows the size and locations of three charges. What would be

Question

The figure at right shows the size and locations of three charges. What would be the net flux through a cube 5m on each edge centered at the origin? (epsilon_0 = 8.85 times 10^-12 C^2/Nm^2) The square surface shown in Fig. 24-25 measures 3.2 mm on each side. It is immersed in a uniform electric field with magnitude E = 1800 N/C. The field lines make an angle of 35 degree with a normal to the surface, as shown. Take that normal to be directed "outward, " as though the surface were one face of a box. Calculate the electric flux through the surface.

Explanation / Answer

The flux is just the charge inside the cube divided by Eo
Qx = 3 × 10-6 C / 8.85 × 10-12C2/N-m2
= 338.9 kN-m2

At Qy = 5 * 10-6 / 8.85 * 10-12

= 564.9 kN-m2

At Qy = 2 * 10-6 / 8.85 * 10-12

= 225.9kN-m2

Therefore the total flux is E = Qx + Qy +Qz

E = 1129.7 kN-m2

2.  The Flux is given by Ex = integration E * A
, where the integral disappears since this is a
constant electric field and we have flat surfaces.

The vector A points upward in the diagram. Thus we have that the flux is just given by
Integral EA cos(35) = 1800N/C ×32.768 * 10-9 m2 × cos(35) = 4.8 * 10-5 N-m2/C

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