2.1) Thunder Clouds and Lightning A thunder cloud with a projected area of A = 5

Question

2.1) Thunder Clouds and Lightning A thunder cloud with a projected area of A = 50 km2 at an altitude of h = 1.5 km above the ground discharges through lightning bolts, each lasting 0.5 ms. The breakdown electric field strength for large distances in humid air is Ea = 1000mm. Hint: The magnitude of the surface charge density on the ground is equal to that of the cloud. 2.1.1) (4) Calculate the total charge Q and the surface charge density = Q/A of the cloud. 2.1.2) (4) Calculate the average bolt current Ib = Q/t when the cloud completely discharges with n = 10 bolts, what is the average power per bolt? How large is the total energy released by the cloud? 2.2) The Millikan Oil-Drop Experiment 2.2.1) (5) An oi drop carries six negative elementary charges, has a mass of 1.6 10-12 g, and falls with a terminal velocity (constant) in air. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward? Start with a vector equation and a sketch.

Explanation / Answer

The oil drop falls with a terminal velocity in air.

so net force acting on the oil drop=0

As weight of oil drop always acts in the downward direction,

so downward force = weight of oil drop = mass of oil drop x g

= 1.6 x 10-12 g x 9.8 m/s2

= 1.6 x 10-15 kg x 9.8 m/s2

= 15.68 x 10-15 N

The vertical upward force will be due to electric field to cancel out the downward force

So, vertical upward force = force due to electric field

= charge on the oil drop x Electric field

= 6 x 1.6 x 10-19 C x Electric field = 15.68 x 10-15 N

So, Electric field = 16333.33 N/C

ALL THE BEST IN THE COURSE WORK

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