# Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.74 * 10^4 N/C as shown in the figure below.

Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.74  104 N/C as shown in the figure below.

(a) Calculate the electric flux through the vertical rectangular surface of the box.
kN · m2/C

(b) Calculate the electric flux through the slanted surface of the box.
kN · m2/C

(c) Calculate the electric flux through the entire surface of the box.
kN · m2/C
asked Feb 10, 2012 in Physics

The formula for electric flux is as follows:

Electric flux = E*A (* E = electric field, A = area)

and if electric flux is at an angle from the normal (which is perpendicular) the formula is:

Electric flux = E*A*cos(theta)   (* E = electric field, A = area, theta = angle between the normal of E and A)

For part A: First find the area of the rectangular surface.

Area of rectangle = length * width

A = 30cm * 10cm = 300 cm^2 = 0.03 m^2

and the electric field is given as

E =   6.74  104 N/C

Now, notice that since the electric field is entering the surface, it is at an angle of 180 degrees with the surface.

So plugging into the formula we get:

Electric flux = E*A*cos(theta)

Electric flux = (6.74  104 N/C) * (0.03 m^2 )*cos(180)  =  -2022 N · m2/C

Electric flux = -2022 N · m2/C  =  -2.022 kN · m2/C

answered Feb 14, 2012 by ~Expert~ (3,020 points)
selected Feb 15, 2012 by awesome
Part B can solved similarly by first finding the area of the slanted surface.

This area is equal to,

A = length * width = 30cm * width.

A = 30cm * (10cm / cos(60)) = 600 cm^2  = 0.06 m^2

Plugging into the formula we get:

Electric flux = (6.74  104 N/C) * (0.06 m^2 )*cos(60)  =  2022 N · m2/C

Electric flux =  2022 N · m2/C  = 2.022 kN · m2/C

Part C, the elctric flux through the entire surface would just be the sum of part A and B,

whic is   - 2.022 kN · m2/C + 2.022 kN · m2/C  = 0

so electric flux = 0