Answer:
Ф1=295.96Nm^2/C
Ф2=469.73Nm^2/C
Explanation:
The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 1.70 m², while surface (2) has an area of 4.00 m². The electric field in the drawing is uniform and has a magnitude of 210 N/C. Find the magnitude of the electric flux through surface (1) if the angle θ made between the electric field with surface (2) is 34.0
we have two surfaces where we know angle of surface 1 lets call it
s1= 34.
Therefore to find s2
s2 (the angle from surface 2) we have that
s2=180-(90+s1),
so s2=180-(90+34),
thus s2=56 degrees.
Flux equation reads as Φ=ΕΑ,
where Φ is the flux,
E is the electric field and
A is the surface area.
So with respect to the angles and the figure provided,
we have Φ=EAcos(s).
So we can solve further by writing
. For Surface 1 we have
Φ1=EAcos(s1)=210 x 1.7 x cos 34,
so Φ1=295.96,
approximately to an whole number
Φ1=296 Nm^2/c.
Similarly for Φ2, we have
Φ2=EAcos(s2)
=210 x 4 x cos34=469.7,
thus Φ2=469.7Nm^2/c.
