Answer:
[tex]400000\ \text{N/C}[/tex]
Explanation:
[tex]q_1[/tex] = Charge at 3000 m = 40 C
[tex]q_2[/tex] = Charge at 1000 m = -40 C
[tex]r_1[/tex] = 3000 m
[tex]r_2[/tex] = 1000 m
k = Coulomb constant = [tex]9\times10^9\ \text{Nm}^2/\text{C}^2[/tex]
Electric field due to the charge at 3000 m
[tex]E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}[/tex]
Electric field due to the charge at 1000 m
[tex]E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}[/tex]
Electric field at the aircraft is [tex]E_1+E_2=40000+360000=400000\ \text{N/C}[/tex].