Electric charge can accumulate on an airplane in flight. You may have observed n

Question

Electric charge can accumulate on an airplane in flight. You may have observed needle-shaped metal extensions on the wing tips and tail of an airplane. Their purpose is to allow charge to leak off before much of it accumulates. The electric field around the needle is much larger than the field around the body of the airplane and can become large enough to produce dielectric breakdown of the air, discharging the airplane. To model this process, assume that two charged spherical conductors are connected by a long conducting wire and a charge of 58.0 µC is placed on the combination. One sphere, representing the body of the airplane, has a radius of 6.00 m, and the other, representing the tip of the needle, has a radius of 2.00 cm.

please answer all fields for a & b in the photo. thanks

Electric charge can accumulate on an airplane in flight. You may have observed needle-shaped metal extensions on the wing tips and tail of an airplane. Their purpose is to allow charge to leak off before much of it accumulates. The electric field around the needle is much larger than the field around the body of the airplane and can become large enough to produce dielectric breakdown of the air discharging the airplane. To model this process, assume that two charged spherical conductors are connected by a long conducting wire and a charge of 58.0 HC is placed on the combination One sphere, representing the body of the airplane, has a radius of 6.00 m, and the other, representing the tip of the needle, has a radius of 2.00 cm. (a) What is the electric potential of each sphere? Your response ers from the correct answer by more Than 10" Double check your calculations V Your response differs from the enmect answer by more than Double checkyour cacuations v (b) What is the electric field at the surface of each sphere? 600m magnitude Your response dillers from the comect answer by more than 10". Double check your calculations vim direction from the sphere direction away from the sphere Submit Answer Save Progress Practice Another versco

Explanation / Answer

let r1 = 6 m

r2 = 2 cm = 0.02 m

Q = 58 micro C

let q1 and q2 are the charges on each sphere.

q1 + q2 = 58 micro C -------(1)

in the equilibriu, V1 = V2

k*q1/r1 = k*q2/r2

q1/6 = q2/0.02

q1/q2 = 6/0.02

q1/q2 = 300 ---------(2)
on solving equation 1 and 2

we get

q1 = 57.807 micro C

q2 = 0.1927 micro C

so,

a)
V1 = k*q1/r1 = 9*10^9*57.807*10^-6/6 = 8.67*10^4 volts

V2 = k*q2/r2 = 9*10^9*0.1927*10^-6/0.02 = 8.67*10^4 volts

b)
E1 = k*q1/r1^2 = 9*10^9*57.807*10^-6/6^2 = 1.44*10^4 N/C

E2 = k*q2/r2 = 9*10^9*0.1927*10^-6/0.02^2 = 4.35*10^6 N/C

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