An airplane is flying at altitude 2 000[m] through a thundercloud. If +40[C] is concentrated at altitude 3 000[m] , directly above the airplane, and −40[C] is concentrated at altitude 1 000[m] directly below it, what Electric field intensity (magnitude) is at the aircraft?

At the position of the airplane, the electric field will be due to both the positive and negative charges and will be downwards due to both the charges.

The position is equidistant from both the charges.

We need to calculate the resultant electric field

The contribution due to the positive charge at 3000 m altitude

Using formula of electric field

[tex]E_{+}=k\dfrac{|q|}{r^2}[/tex]

Put the value into the formula

[tex]E_{+}=8.99\times10^{9}\times\dfrac{40}{(3000-2000)^2}[/tex]

[tex]E_{+}=3.59\times10^{5}\ N/C[/tex]

The contribution due to the negative charge at 1000 m altitude

Using formula of electric field

[tex]E_{-}=8.99\times10^{9}\times\dfrac{40}{(2000-1000)^2}[/tex]

[tex]E_{-}=3.59\times10^{5}\ N/C[/tex]

We need to calculate the electric field intensity at the aircraft

The resultant electric field is

[tex]E=E_{+}+E_{-}[/tex]

Put the value into the formula

[tex]E=3.59\times10^{5}+3.59\times10^{5}[/tex]

[tex]E=7.18\times10^{5}\ N/C[/tex]

Hence, The electric field intensity at the aircraft is [tex]7.18\times10^{5}\ N/C[/tex]