Here's the problem:(adsbygoogle = window.adsbygoogle || []).push({});

The passage of a thundercloud overhead caused the vertical electric field strength in the atmosphere, measured at the ground, to rise to 0.1 statvolt/cm [E_0].

(a) How much charge didthe thundercloud contain, in esu per cm^2 of horizontal area?

(b) Suppose there was enough water in the thundercloud in the form of 1-millimeter-diameter [convert to cm and divide by two, call this r] drops to make 0.25 cm of rainfall [d], and that it was those drops which carried the charge. How large was the electric field strength at the surface of one of those drops?

Here's what I got:

(a) E_0 = 2πσ, from which we get σ = E_0/(2π)

(b) Here's where I am not sure whether I am doing the right thing or not.

Consider some part of the water at the ground of area A at the top (assuming level water).

Φ = 4πq and Φ = E_0 * A => q = AE_0/(4π)

This is the total charge there is.

On the other hand,

Adρ = AE_0/(4π) <=> ρ = E_0/(4πd)

Then the charge which is carried by one drop is:

Q = (4/3)πρr^3 => Q = (r^3)E_0/(3d)

Since the field at the surface is Q/R^2, we get:

E = R*E_0/(3d)

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# Homework Help: Tell me whether I solved this electrostatics problem correctly

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