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 Woodlake 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 to charge spherical conductors are connected by a long conducting wire and a charge of 78 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. (a) what is the electric potential of each sphere? r = 6.00 m: V r = 2.00 cm. V (b) what is the electric field at the surface of each sphere? r = 6.00 m: magnitude V/m direction—Select— away from the sphere toward the sphere r = 2.00 cm: magnitude V/m direction—Select— away from the sphere toward the sphere 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 Woodlake 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 to charge spherical conductors are connected by a long conducting wire and a charge of 78 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. (a) what is the electric potential of each sphere? r = 6.00 m: V r = 2.00 cm. V (b) what is the electric field at the surface of each sphere? r = 6.00 m: magnitude V/m direction—Select— away from the sphere toward the sphere r = 2.00 cm: magnitude V/m direction—Select— away from the sphere toward the sphere

Explanation / Answer

given

q = 78 uC

R = 6 m

r = 2 cm

two conducting spheres, of the given radii will have same poteneial as they are connected

now

let q1 and q2 be charge on both of them

then

q1 + q2 = q

also

V = q1/4*pi*epsilon*R = q2/4*pi*epsilon*r

hence

q1 = q2(R/r) = 300*q2

301q2 = q

q2 = q/301

q1 = 300q/301

V = 300*(78*10^-6)*8.98*10^9/6*301

a. V = 116352.15946843 V

b. E1 = kq1/R^2 = 8.98*10^9*300*78*10^-6/36*301 = 19392.026578 V/m

E2 = kq2/r^2 = 8.98*10^9*78*10^-6/301*0.02^2 = 5817607.97342 V/m

direction of both electric fields is away from the sphere

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.