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Question

If anyone can help me out with this I would really appreciate it. I can not come up with the right answer that's in the book. Thank you :D

An airplane is flying through a thunderstorm at a height of 2100 m. (This is a dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If there are charge concentrations of +40.0 C at a height of 3700 m withiin the cloud and -40.0 C at a height of 1000 m, what is the electric field E at the aircraft? (Give the magnitude and direction.)

Answer 1

You have to calculate --both-- fields from both charges and find their vector sum. The field strength (from Coulombs Law) is
E = kq/r² where k = 9x10^9 nm²/C² so the charge of -40C at 1100 below the plane causes a field of
E = 9x10^9 *(-40)/1100² = -2.975x10^5 V

The +40C charge at a distance of 1600 m is
E = 9x10^9 *(40)/1600² = 1.406x10^5 V so the total field is
E = 1.406x10^5 - (-2.975x10^5) = 4.381x10^5 V and it is pointing 'upwards' (from negative to positive)

BTW, I have a bit over 4,000 hours on a commercial pilots license and the real danger in a thunderhead is turbulence and shear. But being struck by lightning in an all metal aircraft is no big deal. It's happened to me many times. It does get real bright and blows out your vision for a couple seconds, and (if the aircraft wiring isn't properly bonded) it can sometimes cause you to lose a few instruments, but it's not really all that dangerous ☺

Doug

Answer 2

You Have All The Parameters There Reqd. Dist And Charge
Use Coulombs Law