By definition, the flux is given as follows:
Φ=EScosθ where E=149N/C is the field (should be constat), S=4πd2 is the surface area (in this case circle with diameter d=430mm=0.43m), and θ is the angle (from the normal to the surface) between field and surface. In the first case θ=0°, since the net is held horizontaly.
Thus, obtain:
Φ=EScosθ=4πEd2≈21.6CNm2 In the second case the angle become θ=90°, then cosθ=0 and the flux is 0.
Answer. 1) 21.6CNm2, 2) .
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