Answer:
The electric flux through the car's bottom is
![1.75* 10^(4) Wb](https://img.qammunity.org/2020/formulas/physics/college/ptg4fccapwzyvyos3zoffqgbz5ql5b7kwf.png)
Solution:
As per the question:
Magnitude of vertical Electric field,
![E_(v) = 1.85* 10^(4) N/C](https://img.qammunity.org/2020/formulas/physics/college/uilhsiteq92jq009oe19vpabmau81p1jkh.png)
The area of the rectangular surface of the car,
![A_(bottom) = 6* 3 = 18 m^(2)](https://img.qammunity.org/2020/formulas/physics/college/2smqcy66jwv32blsgu3s1q2l5062af1fj5.png)
Downward slope at an angle,
![\angle\theta = 19.0^(\circ)](https://img.qammunity.org/2020/formulas/physics/college/yzmgg9st0f7jw6e0kpv3yz61yjkz4l6na2.png)
Now, the electric flux,
is given by:
![\phi_(E) = \vec{E_(v)}.\vec{A_(bottom)} = E_(v)A_(bottom)cos\theta](https://img.qammunity.org/2020/formulas/physics/college/yzgqk7800gsthwfytu8cusar51nyelxk3n.png)
Now, substituting the appropriate values in the above formula:
![\phi_(E) = 1.85* 10^(4)* 18cos19.0^(\circ)](https://img.qammunity.org/2020/formulas/physics/college/l0xn7z3zurmohv87err0z9w95bsr9vqh2g.png)
![\phi_(E) = 1.75* 10^(4) Wb](https://img.qammunity.org/2020/formulas/physics/college/gsp9snm5lb0cyetgwcbs1lvzb2tplz32gi.png)