Find the flux through the equilateral triangle with corners at

[apebeast]

Homework Statement

Find the flux through the equilateral triangle with corners at the points (1m,0,0), (0,1m,0), and (0,0,1m) in x,y,z space (measured in meters) for an electric field with magnitude E=6N/C pointing

• (a) in the z direction,
• (b) parallel to the line y = x.

Homework Equations

Flux = EA;
Surface area of an equilateral triangle = (sqrt(3)/4)(a^2), where a = side of the triangle
Distance formula = sqrt(((x2-x1)^2) + (y2-y1)^2)

The Attempt at a Solution

I approached this using the equation:

Flux = EA, where "E" is 6 N/C.

I tried solving for A by using the formula for the equilateral triangle area: ((sqrt(3)*s^2)/4), where s
is the side of the triangle. I solved for the side using the distance formula.

Now, with all things in consideration, I plugged everything in:

E=6
A=((sqrt(3)*(sqrt(2))^2)/4); s=sqrt(2);

E=6
A=(sqrt(3)/2) or .867

Flux = 3sqrt(3) or 5.1962

Now, I put this answer for question (a), and still got the wrong answer. What am I doing incorrectly?

Thank you kindly for the help.


Nicu

 

[TSny]

Hello, and welcome to PF!

Homework Equations

Flux = EA;

This equation is valid only if the electric field is oriented perpendicular to the area surface. You should have covered how this equation is modified to handle other cases.

 

[champion19]

Find the area vector of the triangle. Remember that it should be normal to the triangle's surface and have magnitude equal to the triangle's area.

 

[Quantum Braket]

Go outside on a sunny day and hold your textbook out and look down at its shadow. How can you orientate the book to maximize the area of the shadow? To minimize it?
Can you think of a mathematical relation that will allow you to calculate the area of the shadow given the orientation of the book wrt the sun?
How does the area of the shadow relate to the flux through a surface as in your problem?

 

[HalfLostAndSearching]

, your approach is correct but your calculations are slightly off. Let's go through the steps again to find the flux through the equilateral triangle.

1. First, we need to find the area of the equilateral triangle. The formula for the area of an equilateral triangle is A = (sqrt(3)/4)*a^2, where a is the side of the triangle. In this case, the sides of the triangle are all 1m, so a = 1m.

2. Next, we need to find the distance from the point (0,0,0) to each of the corners of the triangle. Using the distance formula, we get:

Distance from (0,0,0) to (1,0,0) = 1m
Distance from (0,0,0) to (0,1,0) = 1m
Distance from (0,0,0) to (0,0,1) = 1m

3. Now, we can plug these values into the equation for flux, Flux = EA. We know that E = 6N/C, so we have:

(a) Flux = (6N/C)*(1m)^2*(sqrt(3)/4) = (3sqrt(3)) N*m^2/C

(b) To find the flux with the electric field parallel to the line y = x, we need to find the component of the electric field that is parallel to the surface of the triangle. This can be done by finding the dot product between the electric field vector and a vector normal to the surface of the triangle. In this case, the normal vector is (1,1,1). So, the dot product is:

E dot n = (6N/C)*(1m,1m,1m) dot (1,1,1) = 6N/C

Now, we can find the flux:

Flux = (6N/C)*(1m)^2*(sqrt(3)/4) = (3*sqrt(3)) N*m^2/C

So, the answer for both parts (a) and (b) is Flux = (3*sqrt(3)) N*m^2/C. I hope this helps!