Calculating Ambient Electric Fields in Thunderclouds for Aircraft Safety

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Homework Help Overview

The discussion revolves around calculating the ambient electric field experienced by an aircraft flying through a thundercloud, given specific charge concentrations at different altitudes. The subject area includes concepts from electrostatics and electric fields.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply Coulomb's law to calculate the electric field, but questions whether they are using the correct approach and if they have missed steps. Some participants suggest converting units and question the magnitude of the charge values used.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on unit conversion and questioning the assumptions made about charge values. There is a lack of consensus on the correctness of the calculations, and some participants express uncertainty about their understanding of the concepts involved.

Contextual Notes

There are indications of confusion regarding the values of charge and the appropriateness of the calculations, as well as the relevance of the constant k in the context of the problem.

Binkey24
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Sorry if i have posted under the wrong catagorie physics is new to me so I am still getting my head around it.

Here is the question that is giving me a bit of bother;
An airplane is flying through a thunder cloud at an altitude of 2km. If the cloud has a charge concentration of 40C at an altitude of 3km above the ground and -40C at 1km altitude. What is the ambient electric field exsperienced by the air craft?
Additional information k = 9 x 10^9 N-m/C^2)

so far i have think the equations to be used are F=kQ,Q2 / r^2

this is what i get when i fill the following information = (9*10^9)*(40)*(-40) / 1^2

this equaling -1.44^13
Am i going the right way about it or have i missed a few steps? any guidance would greatly be appreciated.
 
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Convert kilometer to meter.
 
After changing the km into meters

(9*10^9) * (40) * (-40) / 1000^2

this gave me answer of = - 14400000
this answer seems a little big. or is it meant to be this large?
 
Hmm 40C is wayyy too big. But anyways, your not answering the question. What is the electric field experienced by the plane question mark. (Sorry keyboard problems)
 
Not sure if I can help at all, I'm stuck in the same place. I think we are studdying the same thing.
 
Just found this, hope it helps:

See attachment, it didn't want to past in here.

Edit: Note the diference in k from your original question.
 

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