Answer:
[tex]\Phi_E=21\frac{N\cdot m^2}{C}[/tex]
Explanation:
The electric flux through an area is defined as the electric field multiplied by the area of the projected surface on a plane perpendicular to the field.
[tex]\Phi_E=EA(1)[/tex]
According to Gauss law, since the electric filed is uniform, there are no charges in the basket, we obtain the total flux through the sides and bottom, replacing the given values in (1):
[tex]\Phi_E=300\frac{N}{C}\pi(0.15m)^2\\\Phi_E=21\frac{N\cdot m^2}{C}[/tex]
We have that for the Question "A cylindrical wastebasket with a 0.15 m radius opening is in a uniform electric field of 300 N/C, perpendicular to the opening. The total flux through the sides and bottom is:" it can be said that total flux through the sides and bottom is
[tex]Total\ Flux =21.195Nm/C\\\\[/tex]
From the question we are told
A cylindrical wastebasket with a 0.15 m radius opening is in a uniform electric field of 300 N/C, perpendicular to the opening. The total flux through the sides and bottom is:
Generally the equation for the flux is mathematically given as
[tex]\phi=E(\pir^2)cos90[/tex]
Therefore
[tex]Total\ Flux =E\pir^2\\\\Total\ Flux =(300)(3.14)(0.15)^2\\\\Total\ Flux =21.195Nm/C\\\\[/tex]
Therefore
total flux through the sides and bottom is
[tex]Total\ Flux =21.195Nm/C\\\\[/tex]
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