The vertical electric below a thundercloud is about 10^4 Vm^-1.
The charge will carry 4.188 X 10^-9 C when the radius of a raindrop is1mm.
(1) If the breakdown strength of the air is approximately 3MVm^-1, determine whether it is likely that the drop can carry such a charge.
(2) The drop now splits into two identical smaller drops so that each carries the same amount of charge. Calculate the new electic field strength at the surface of the drops.
(density of water = 1000 kgm^-3, ε = 8.9 X 10^-12 N^-1m^-2C^2)
- 天同Lv 71 decade agoFavorite Answer
1. Electric field intensity E = k.Q/R^2
where k is a constant (=9x10^9 N-m^2/C^2 )
Q is the charge
and R is the radius of the charge
Hence, electric field intensity at the surface of the rain drop
= 9x10^9 x 4.188x10^-9/0.001^2 v/m = 3.77x10^7 v/m = 37.7 Mv/m
which has already exceeded the breakdown electric field intensity
2. The new charge on each drop = Q/2
New radius r = [cube-root(2)].R
hence , new electric field intensity E' is
E' = k.(Q/2)/2^(2/3).R^2 = 2^-(5/3).kQ/R^2 = 2^(-5/3).E = 0.3E
= 0.3 x 37.7 Mv/m = 11.3 Mv/m