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 13.0 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
Direction (toward the sphere or away from the sphere)__________
r = 2.00 cm
______________V/m
Direction (toward the sphere or away from the sphere)

Explanation / Answer

A) The radius of airplane r_1 = 6 *10^- 2 m The radius of needle r_2    = 2 * 10^- 2 m we have q_1 / q_2 = r_1 / r_2     or q_1 = 3 q_2    q_1 + q_2 = 13 µC       3 q_2 +q_2 = 13*10^-6C               q_2 = 3.25 *10-6C      &    q1 = 9.75 *10-6C              thepotential of the air plane is               V_1 = k q_1 / r_1           = 9 *10^9 * 9.75 *10^-6 C / 6*10^-2 m            = 1.46 * 10^6 V The potential of needle V_2    = k q_2 / r_2                                                                                               = 9 *10^9 * 3.25 *10^-6 C /  2.0*10^-2 m                                                  = 1.46 * 10^6 V                                                  = 1.46 * 10^6 V B)                 the electricfield of the airplane is                                  E = k q_1 / ( r_1 )^2                                   E_1 = 9 *10^9 * 9.75 *10^-6 C / (6*10^-2 m )^2                                            = 2.4 *10^7 V/m                                            = 2.4 *10^7 V/m


The electric field of the needle is                              E_2    = k q_2 /( r_2 )^2                                                                              = 9 *10^9 * 3.25 *10^-6 C /  (2.0*10^-2 m )^2                                         = 7.31 *10^7 V/m                                                                              = 9 *10^9 * 3.25 *10^-6 C /  (2.0*10^-2 m )^2                                         = 7.31 *10^7 V/m                                                                           

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