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 extenslons on the wing tips and tall of an alrplane. Thelr purpose is to allow charge to leak off before much of it accumulates. The electric field around the needle is much larger than the fleld around the body of the airplane and can become large enough to produce dielectric breakdown of the alr, discharging the airplane. To model this process, assume that two charged spherical conductors are connected by a long conducting wire and a charge of 74.0 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? r= 6.00 m: r= 2.00 cm: Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. V (b) What is the electric field at the surface of each sphere? r-6.00 m: magnitude V/m direction -Select- r-2.00 cm magnitude V/m direction Select

Explanation / Answer

a)
Both spheres must be at the same potential
Ke*q1/r1 = Ke*q2/r2

also
q1+q2 = q = 74 uC

Solving for q1 in equation 1 in plugging into 2,
q2*r1/r2 + q2 = 74*10^-6 C

q2(6/2 + 1) = 74*10^-6

q2 = (74*10^-6)/4
= 18.5*10^-6 C

Then q1 = q-q2
q1 = 74*10^-6 - 18.5*10^-6
= 55.5*10^-6 C
= 55.5 uC

V1 = ke*q1/r1
V1 = V2
=> V1 = (9e9 * 55.5*10^-6)/(6e-2)
= 8.325*10^6 V

b)
E = kq/r^2

E1 = 8.325*10^6/6e-2
= 1.38*10^8 V/m .... away

E2 = 8.325*10^6/2e-2
= 4.16*10^8 V/m .... away

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