PHYSIC 難題21

Electric field between the base of the cloud and ground
= 200x10^6/500 v/m = 4 x 10^5 v/m
Hence, electric force on the raindrop
= (4x10^-12) x (4x10^5) N
= 1.6x10^-6 N

The answer is option A

Let consider the base of the cloud and the ground as two metal plates of a capacitor.

The following equations should be taken into account here:
1. V = Ed where V = p.d. between the two surfaces, E = electric field between the two surfaces and d = distance of the cloud's base from the ground
2. F = qE where F = electric force, q = electric charge of the raindrop and E = electric field between the two surfaces

Calculations:
1. Electric field: E = V/d = 200 000 000 / 500 = 400 000 V/m
2. Electric force: F = qE = (0.000000000004)(400 000) = 0.0000016 N

Note that unit of q is C and that of E = V/m = W/(Am) = Ws/(Cm). Since Power = Force x velocity, unit of power = N m/s and thus unit of E = (N m/s)s/(Cm) = N/C. Thus, unit of qE is C x N/C = N which is equivalent to unit of F.

2012-05-12 19:14:50 補充：
Option A is correct