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 72.0 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.

(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

r = 2.00 cm: magnitude______ V/m

Explanation / Answer

The potential on both spheres will be same.

let r1 and r2 are the radii of the spheres.

r1 = 6 m

r2 = 2 cm = 0.02 m

q1 + q2 = 72 micro C ---(1)

V1 = V2

k*q1/r1 = k*q2/r2

q1/r1 = q2/r2

q1/6 = q2/0.02

q1 = 300*q2

from equation 1

300*q2 + q2 = 72*10^-6 C

q2 = 72*10^-6/301

= 0.239*10^-6 C

so, q1 = 72*10^-6 - 0.239*10^-6

= 71.761*10^-6 C

a) potentail on larger sphere, V = k*q1/r1

= 9*10^9*71.761*10^-6/6

= 1.076*10^5 volts

potential on smaller sphere = 1.076*10^5 volts

b) Electric field on larger sphere = k*q1/r1^2

= 9*10^9*71.761*10^-6/6^2

= 1.794*10^4 V/m

Electric field on larger sphere = k*q2/r2^2

= 9*10^9*0.239*10^-6/0.02^2

= 5.38*10^6 V/m

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