Airplane electric field

  • #1
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Homework Statement


An airplane is flying through a thunderstorm at a height of 1800m. If there are charge concentrations of 40 C at height 3400m within the cloud and -28 C at height 772m, what is the strength of the electric feild, E, at the aircraft? Answer in V/m


Homework Equations


E = F/ q


The Attempt at a Solution


I drew a diagram with distances and charges. I know how to find the force between the -28 charge and 40 charge by using coulomb's equation. but what do i plug in for "q" (the charge) in the E=F/q

or is this approach wrong all together?
 

Answers and Replies

  • #2
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so i thought about it further..

I calculated force using Coulomb's equation and got 1.4579x10^6 N

then using E=F/q i calculated the electric field at each of teh charges (-28 and 40).
Electric field at charge of -28 is 5.20678x10^4
electric field at charge of 40 is 3.64474 x 10 ^4 ..
now how do I use this to get the electric field at the aricraft? Please help me out. Thanks!
 
  • #3
LowlyPion
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The electrical field can be given by kq/r2

The presumption is that the airplane is influenced by charges directly above and below.

To find the total field at the airplane you simply use superposition of the fields at that point. But mind your signs and directions, because the electric field is a vector field.
 
  • #4
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so since one of the charges is negative , does that mean one of the lectric fields will be negative?
 
  • #5
rl.bhat
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First of all find the distances between positive charge and airplane and negative charge and airplane. Then using the formula provided by LowlyPion in post #3, calculate electric fields on the airplane.
At any point electric field is away from positive charge and towards negative charge.
Now try.
 
  • #6
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ok so I did all of that. Just to check, Can electric fields be negative?
 
  • #7
gabbagabbahey
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ok so I did all of that. Just to check, Can electric fields be negative?

Yes; the negative sign indicates that the field points in the opposite direction.

For example, an electric field E=-2(N/C) j would have a magnitude |E|=2(N/C) and point in the negative y-direction.
 
  • #8
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ok that makes sense. Thank you. One more thing, the answer has to be in units of V/m
but the way I worked it out I got N/C units. how would i make the conversion?
 
  • #9
gabbagabbahey
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1V=1Nm/C

so

1N/C= 1V/m :smile:
 
  • #11
gabbagabbahey
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Welcome :smile:
 
  • #12
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unfortunately my answer is still incorrect =(

so for the 3400 m and a charge of +40 , in relation to the aircraft the electric field i calculated was 140469 N/C
And for 772m and a charge of -28 , my electric field was -238194 N/C
So then I added those two and got -97725.6 N/C.
The hw program told me it was "wrong". Please help =(
 
  • #13
gabbagabbahey
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Using k=8.9876 x 10^9 N/C^2 I get -97699.5 V/m. So it might just be that you are supposed to use a more accurate value for k. Or you might be supposed to write your answer without the negative sign, since the sign indicates the direction of the field rather than the strength of it.
 
  • #14
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they gave the value of k=8.99 x 10^9 Nm^2/C^2. I tried my answer without the sign and it still said "wrong" =(
 
  • #15
gabbagabbahey
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Did they specify a number of significant digits to use?
 
  • #16
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nope. Infact they tell you to not use sig figs or rounding
 
  • #17
gabbagabbahey
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In that case, I;m not sure where the problem is. I get the same answer as you.

Is the question written word for word as in your first post?
 
  • #18
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Yes, same wording. the only thing i left out was their given value of coulomb's constant. but i mentioned it in another post..I have no idea what to do =(
 
  • #19
gabbagabbahey
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Hmmm...strange. I'd ask your prof or TA on this one. Sometimes automated systems are buggy.
 
  • #20
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Yes, I hope that is the case. Anyhow, Thank you very much once again! =]
 
  • #21
rl.bhat
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[QUOTE=itryphysics;2044009]unfortunately my answer is still incorrect =(

so for the 3400 m and a charge of +40 , in relation to the aircraft the electric field i calculated was 140469 N/C
And for 772m and a charge of -28 , my electric field was -238194 N/C
So then I added those two and got -97725.6 N/C.
The hw program told me it was "wrong". Please help =([/QUOTE]


In the first case the electric field is from +chargs to airplane and in the second case the field is from airplane to negative charge. So they are in the same direction. Therefore you have to add two fields with out taking into consideration their signs, to get the resultant field.
 
  • #22
gabbagabbahey
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In the first case the electric field is from +chargs to airplane and in the second case the field is from airplane to negative charge. So they are in the same direction. Therefore you have to add two fields with out taking into consideration their signs, to get the resultant field.

You're right; I can't believe I missed that! :redface:
 
  • #23
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Thank you SO much! =]
 

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