Solving Electric Field Strength at Aircraft in Cloud: 1.03357e8 + 2.64009e9

[Spectre32]

AN airplane is flying through a cloud. The airplane is at 2100M and the top height of the cloud is 4100M @ 23 C. The bottem of the cloud is 500m @-47 C. What is the strength of the electric field E at the aircraft.

Now I broke it up into to parts, for the top of the cloud and the bottem.

Top of the cloud:

Using coulumbs law: 8.98755 x 10^9* 23/ 4100-2100) = 1.03357e8

Bottem of the cloud:

8.98775 x 10^9 * 47/ 2100-500 = 2.64009e9

I added both of those up and i submited my answer and it said it was wrong. Can anyone aid me?

 

[chroot]

The distance should be squared.

Also, make sure you put your parentheses in the right place. I assume that you are actually typing the numbers into your calculator correctly, but just be careful.

8.98775 * 10^9 * 47 / 2100 - 500

is not the same as

(8.98775 * 10^9 * 47) / (2100 - 500)

- Warren

 

[Doc Al]

Originally posted by Spectre32

I added both of those up and i submited my answer and it said it was wrong. Can anyone aid me?

Assuming that the charged clouds can be treated as point sources of the electric field: use $E=\frac{kq}{r^2}$. (You forgot to square your distances.)

 

[Spectre32]

crap... wow I'm so retarded thanks a lot doc