1) What is the approximate magnitude of the electric field between the charge ce

Question

1) What is the approximate magnitude of the electric field between the charge center and the ground?

A) 4 x 10^4 V/m                B) 4 x 10^5 V/m

C) 4 x 10^6 V/m                D) 4 x 10^7 V/m

A) B) C) D) Storm clouds build up large negative charges, as described in the chapter. The charges dwell in charge centers, regions of concentrated charge. Suppose a cloud has -25 C in a 1.0-km-diameter spherical charge center located 10 km above the ground, as sketched in the figure. The negative charge center attracts a similar amount of positive charge that is spread out on the ground below the cloud. The charge center and the ground function as a charged capacitor, with a potential difference of approximately 4 x 10^8 m The large electric field between these two ^'electrodes^' may ionize the air, leading to a conducting path between the cloud and the ground. Charges will flow along this conducting path, causing a discharge of the capacitor - a lightning strike. 1) What is the approximate magnitude of the electric field between the charge center and the ground? A) 4 x 10^4 V/m B) 4 x 10^5 V/m C) 4 x 10^6 V/m D) 4 x 10^7 V/m 2) What is the approximate capacitance of the charge center-ground system? 3) If the cloud transfers all of its charge to the ground via several rapid lightning flashes lasting a total of 1.90 s, what is the average power?

Explanation / Answer

1 ) Electric field between the charge and the center is                  E = V / d potential difference of a charged capacitor is V = 4*10^8 V distance d = 10 km                  E = 4*10^8 V / 10 *10^3 m                        = 4*10^4 V /m 2 ) Capacitance of the charge center ground system is                charge is q = -25 C          Capacitor is C = Q / V                                     = 25 C / 4*10^8 V                                      = 6.25*10^-8 F 3 ) Energy stored in a capcitor is E = 1/2 CV^2                         = 1/2 * 6*10^-8 F * ( 4*10^8)^2                        = 50*10^8 J           Average Power is P = E / t                                           = 50*10^8 J / 1.9 s                                            = 2.7*10^9 W                                             = 2.7 GW

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