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 77.0 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

direction

r = 2.00 cm: magnitude V/m

Explanation / Answer

let charge on the larger sphere be Q, and on the smaller one be q
then given Q + q = 77 micro C

Now, both the spheres are connected so they are at the same potential V

now capacitance of a sphere is given by C = 4*pi*epsilon*r [ where epsilon is permittivity of free space and r is radius of the sphere]
given radii
larger sphere R = 6 m
smaller sphere, r = 0.02 m

so, capacitances are C1 = 4*pi*epsilon*R1, C2 = 4*pi*epsilon*R2

so using Q = CV
we get
q = C2*V
Q = C1*V

q + Q = (C1 + C2)V
V = (q + Q)/(C1 + C2) = 77*10^-6*8.98*10^9/(6.02) = 114860.4651 V is the potential of both the spheres

b. V on surface of sphere is V = kq/r
   so E = kq/r^2 = V/r

   so E1 = V/R = 19143.41 V/m
   E2 = V/r = 5743023.2558 V/m

direction is radially outward from the sphere

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