Question

The figure shows the charge distribution within a thundercloud.
There is a charge of 40 C at a height of 10 km, -40 C at 5 km,
and 10 C at 2 km. Considering these charges to be pointlike,
find the net electric field produced at a height of 8 km and a
horizontal distance of 3 km.

Answer 1

As $\vec{E}=-grad\phi$ hence $E=\sqrt{{(\frac{\partial\phi}{\partial x})^2}+(\frac{\partial\phi}{\partial x})^2}$ where $\phi$ is field potential at point $(x,y)$

As $\phi=\phi_1+\phi_2+\phi_3=1/4\pi\epsilon_0(\frac{Q_1}{r_1}+\frac{Q_2}{r_2}+\frac{Q_3}{r_3})$ where

$r_1=\sqrt{x^2+(y-2)^2}, r_2=\sqrt{x^2+(y-5)^2}, r_3=\sqrt{x^2+(y-10)^2}$ and

$Q_1=10C, Q_2=-40C, Q_3=40C$ then

$(\frac{\partial\phi}{\partial x})_{(3,8)}\approx-1.0\sdot10^4V/m, (\frac{\partial\phi}{\partial x})_{(3,8)}\approx-0.14\sdot10^4V/m$ then $E\approx1.01\sdot10^4V/m$