Question

Capacitance of a Thundercloud The charge center of a thundercloud, drifting 3.0 km above the earth's surface, contains 20 C of negative charge. Assuming the charge center has a radius of 1.0 km, 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.

Answer 1

Concepts and reason

The concept required to solve the given question is parallel plate capacitor.

Initially, calculate the area of the parallel plate capacitor by using the radius of the charge center. Capacitance of the system is calculated by using the relation between the area of the parallel plate capacitor, capacitance of the capacitor, and distance between the parallel plates. Then, the potential difference between the charge center and ground is calculated by using the charge on the capacitor and capacitance of the capacitor.

The average strength of the electric field between the cloud and ground is calculated by using potential difference between charge center and ground, and the distance between the charge center and Earth’ surface. Finally, the electrical energy stored in the system is calculated by using the capacitance of the capacitor and potential difference between the charge center and ground.

Fundamentals

Capacitance of a parallel plate capacitor filled with air is calculated by using the formula,

$C = \frac{{{\varepsilon _0}A}}{d}$

Here, A is the area of the plates, d is the separation between the plates, and ${\varepsilon _0}$ is the permittivity of space.

The value of permittivity of space is,

${\varepsilon _0} = 8.85 \times {10^{ - 12}}{\rm{ F/m}}$

The charge on a capacitor is calculated by using the formula,

$Q = CV$

Here, C is the capacitance of the capacitor and V is the potential difference between the two plates of the capacitor.

Potential difference between the plates of the capacitor is,

$V = Ed$

Here, E is the electric field strength between the two plates of the parallel plate capacitor and d is the distance between the plates.

Energy stored in a capacitor is calculated by using the formula,

$U = \frac{1}{2}C{V^2}$

Here, C is the capacitance of the capacitor and V is the potential difference between the two plates of the capacitor.

(a)

Calculate the capacitance of the system,

Capacitance of the system is calculated by using the formula,

$C = \frac{{{\varepsilon _0}A}}{d}$

Here, A is the area of the charge center, d is the distance between the charge center and the Earth’s surface, and ${\varepsilon _0}$ is the permittivity of space.

Area of the charge center is,

$A = \pi {r^2}$

Here, r is the radius of the charge center.

Substitute $\pi {r^2}$ for A in equation $C = \frac{{{\varepsilon _0}A}}{d}$ .

$C = \frac{{{\varepsilon _0}\pi {r^2}}}{d}$

Substitute 1.0 km for r, 3.0 km for d, and $8.85 \times {10^{ - 12}}{\rm{ F/m}}$ for ${\varepsilon _0}$ .

$\begin{array}{c}\\C = \frac{{\left( {8.85 \times {{10}^{ - 12}}{\rm{ F/m}}} \right)\pi {{\left( {1.0{\rm{ km}}\left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right)} \right)}^2}}}{{\left( {3.0{\rm{ km}}} \right)\left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right)}}\\\\ = 9.27 \times {10^{ - 9}}{\rm{ F}}\\\end{array}$

(b)

Calculate the potential difference between the charge center and ground.

Potential difference between charge center and ground is,

$V = \frac{Q}{C}$

Here, Q is the charge on the charge center and C is the capacitance of the system.

$\begin{array}{c}\\V = \frac{{20{\rm{ C}}}}{{9.27 \times {{10}^{ - 9}}{\rm{ F}}}}\\\\ = 2.16 \times {10^9}{\rm{ V}}\\\end{array}$

(c)

Calculate the electric field between the cloud and ground.

Electric field between the cloud and the ground is,

$E = \frac{V}{d}$

Here, V is the potential difference between the cloud and ground and d is the distance between the cloud and the earth’s surface.

Substitute $2.16 \times {10^9}{\rm{ V}}$ for V and 3.0 km for d.

$\begin{array}{c}\\E = \frac{{2.16 \times {{10}^9}{\rm{ V}}}}{{3.0{\rm{ km}}\left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right)}}\\\\ = 7.20 \times {10^5}{\rm{ V/m}}\\\end{array}$

(d)

Calculate the electrical energy stored in the system.

Energy stored in the system is,

$U = \frac{1}{2}C{V^2}$

Here, C is the capacitance of the system and V is the potential difference between the cloud and ground.

Substitute $9.27 \times {10^{ - 9}}{\rm{ F}}$ for C and $2.16 \times {10^9}{\rm{ V}}$ for V.

$\begin{array}{c}\\U = \frac{1}{2}\left( {9.27 \times {{10}^{ - 9}}{\rm{ F}}} \right){\left( {2.16 \times {{10}^9}{\rm{ V}}} \right)^2}\\\\ = 2.16 \times {10^{10}}{\rm{ J}}\\\end{array}$

Ans: Part a

The capacitance of the system is $9.27 \times {10^{ - 9}}{\rm{ F}}$ .