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 27.0 µC is placed on the combination. One sphere, representing the body of the airplane, has a radius of 6.00 cm, 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 cm
V
r = 2.00cm
V
(b) What is the electric field at the surface of each sphere?
r = 6.00 cm
V/m 4
r = 2.00 cm
V/m

Explanation / Answer

Since the wire is long, we can assume that the field due to one is not disturbed by other

Total charge Q=q1 + q2 = 27 *10-6 C ------(1)

Since they are connected by a metal wire, they will be having same potential

ie kq1/(r1)2 = kq2/(r2)2 --------(2)

where r1= 6 cm =6*10-2 m and r2= 2 cm= 2*10-2 m

Since q1 + q2 = 27 *10-6 C , q2= 27 *10-6 -q1, substitute this in equation (2)

  kq1/(r1) = k(27 *10-6 -q1)/(r2)

q1/(6*10-2) = (27 *10-6 -q1)/(2*10-2)

q1= 2.03*10-5 C so q2= 27 *10-6 -2.03*10-5 = 6.7*10-6 C

a) electric potential , V =  kq1/(r1) = 9*109* 2.03*10-5 /(6*10-2)

= 3.03 MV

b) electric field r=6 cm

E= kq1/(r1)2 = 9*109* 2.03*10-5 /(6*10-2)2 = 50.75 MV/m

electric field r=2 cm

E= kq2/(r2)2 = 9*109* 6.7*10-6 /(2*10-2)2 = 150.75 MV/m

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