# Electric Flux

1. Oct 31, 2003

### discoverer02

Stumped again!!!

My answer doesn't agree with the answer in the book for this problem:

A pyramid with a 6.00-m-square base and height of 4.00 m is placed in a vertical electric field of 52.0 N/C. Calculate the total electric flux through the pyramid's four slanted surfaces.

Maybe the problem's with my geometry?

[0] = EAcos[the].

E is given. I've calculated the area of one of the faces to be 5.121348 m^2 and [the] = 17.0239 degrees.

The total electric flux through the four slanted sides should be:
4(52.0 N/C)(5.121348 m^2)(cos(17.0239 degrees)) right?

Wrong?

I come up with a total electric flux of 1018 (Nm^2)/C

The book comes up with 1.87 (KNm^2)/C

Please show me where I've gone wrong.

Thanks.

2. Oct 31, 2003

### gnome

LOL

Hi d02.

I look forward to answering your questions because you're generally asking about the same topics I need to review myself, & they're usually quite challenging.

But you're gonna die when you read the answer to this one.

The total flux through the pyramid's sides is the same as the total flux through it's base, right?

52 N/C * 36 m^2 = 1872 N.m^2/C = 1.87 kN.m^2/C

TGIF??

3. Oct 31, 2003

### discoverer02

Hi gnome,

Arghh! You're right. If E's constant it should be the same. I also made the mistake of taking "6.00-m-square base" to mean 6 m^2 area and not realizing that for the [the] I chose (the angle between the vertical and the face of the pyramid) the projection of the area vector on E would be sin[the] and not cos[the].

It all adds up now.

Sometimes my brain just seems to check out on me!!!

Thanks for clearing things up for me yet one more time.