Question

In a thunderstorm, charge is separated through a complicated mechanism that is ultimately powered by the Sun. A simplified model of the charge in a thundercloud represents the positive charge accumulated at the top and the negative charge at the bottom as a pair of points charges.
A. what is the magnitude and direction of the electric field produced by the two point charges at point P, which is just above the surface of the Earth?
B. Thinking of the Earth as a conductor, what sign of charge would accumulate on the surface near point P?

Answer 1

A) Magnitude and direction of electric field at point P will be a vector sum of the two point charges.
Equation of an electric field around a point charge A is:
\[\ \hat{E_A}=\frac{Q}{4\pi\epsilon r^2 }\hat{r} \]

E-electric field, Q - effective charge at point A , epsilon - dielectric permittivity, r- distance from point A, r^(hat) is a unit vector.

Note that Q can be either negative or positive, since there are two types of charge.

So, at the point P you will have: \[\ \hat{E_P} = \hat{E_A} + \hat{E_B}\]
Since E_p is a vector it will determine the magnitude and direction of electric field at point P.

B)Lets call the top of a cloud, which has a net positive charge, point A and lets call the bottom, with net negative charge, point B. Lets also assume that both of these two points have the same charge (only one is negative and the other is positive).
Since electric field degrades with distance, there will be more "effect" from the closer point. that is point B, negative charge.

So, negative is closer and has more effect on the electrons in the conductor (earth). As you know, charges with the same sign repel each other and charges with the opposite sign attract each other. Now, electric field will push down the same signed charges (negative) and the opposite charge (positive) will accumulate at the surface.


You could sketch the whole thing on a piece of paper with field lines and everything and maybe see it more clearly.