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°.

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.