Question

The charge center of a thundercloud, drifting 3.0km above the earth’s surface, contains 20C of negative charge. Assuming the charge center has a radius of 1.0km, and modeling the charge center and the earth’s surface as parallel plates, calculate: (a) the capacitance of the system; (b) the potential difference between charge center and ground; (c) the average strength of the electric field between cloud and ground; (d) the electrical energy stored in the system.

Short answer

(a) $C=9.27nF$

(b)$V_{CG}=2.16GV$

(c)$E=7.2\times {10}^{5}V/m$

(d)$U=21.6GJ$.

Step-by-step solution

Parameters.

$\begin{array}{l}d=3.0\mathrm{km}=3.0\times {10}^{3}m\\ Q=20C\\ r=1km=1\times {10}^{3}m\end{array}$

(a) Calculating the capacitance of the system.

The capacitance is given by

$C=\frac{{\in }_{0}A}{d}$ (1)

All is known except A so consider the cloud to be a circular disk and the area is given by

$A=\pi r^2=3.14\times {10}^6{m}^2$

Substituting in (1) yields

$C=\frac{8.85x{10}^{-12}\times 3.14\times {10}^6}{30\times {10}^3}=9.27\times {10}^{-9}F$.

(b) Calculating the potential difference between the charge center and ground.

The voltage difference between the cloud and the ground is given by

$V_{CG}=\frac{Q}{C}=\frac{20}{9.27\times {10}^{-9}}=2.16\times {10}^{9}V$.

(c) Calculating the average strength of the electric field.

The electric field flows through the air is given by

$EA=\frac{Q}{{\in }_0}\Rightarrow E=\frac{Q}{A{\in }_0}$ (3)

Substituting in (3) yields

$E=\frac{20}{8.85\times {10}^{-12}\times 3.14\times {10}^6}=7.2\times {10}^{5}V/m$

(d) Calculating the electrical energy stored in the system.

The energy stored in system is given buy

$U=\frac{1}{2}QV=\frac{1}{2}\times 20\times 2.16\times {10}^9=2.16\times {10}^{10}J$.