Answer:
The electric flux through the car's bottom is [tex]1.75\times 10^{4} Wb[/tex]
Solution:
As per the question:
Magnitude of vertical Electric field, [tex]E_{v} = 1.85\times 10^{4} N/C[/tex]
The area of the rectangular surface of the car, [tex]A_{bottom} = 6\times 3 = 18 m^{2}[/tex]
Downward slope at an angle, [tex]\angle\theta = 19.0^{\circ}[/tex]
Now, the electric flux, [tex]\phi_{E}[/tex] is given by:
[tex]\phi_{E} = \vec{E_{v}}.\vec{A_{bottom}} = E_{v}A_{bottom}cos\theta[/tex]
Now, substituting the appropriate values in the above formula:
[tex]\phi_{E} = 1.85\times 10^{4}\times 18cos19.0^{\circ}[/tex]
[tex]\phi_{E} = 1.75\times 10^{4} Wb[/tex]
