Calculate Flux in a closed Triangle

[RedPhoenix]

Homework Statement

Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.80 & 104 N/C as shown in Figure P24.4. Calculate the electric flux through (a) the vertical rectangular surface, (b) the slanted surface, and (c) the entire surface of the box.

Homework Equations

Flux = EAcos( )

The Attempt at a Solution

for a) I did

(7.8E4)(.1x.3)cos(0) = 2.34kN

The answer should be negative, why?
Would it be cos(180) instead? What is the reason for this?

for b), the slanted plane, the answer is 2.34kN, but why would it be? I did (7.8E4)(.1x.3)cos(0), which gives the right answer, but why is it 0? not 60 or something?

c) I am assuming because a and b cancel each other when added together.


A little input would be nice, thanks :)

Matt

 

[Donaldos]

a) The surface normal and the electric field point in opposite directions.

b) The area of the slanted surface is not $$0.1 \times 0.3$$ but $$\frac{0.1}{\cos 60} \times 0.3$$

$$\Phi_E=(7.8\times 10^4) \times \frac{0.1}{\cos 60} \times 0.3 \times \cos 60=(7.8\times 10^4) \times \0.1\times 0.3$$