Chapter 4: Q58P (page 812)

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

Expert verified

(a) $C = 9.27 n F$

(b) $V_{C G} = 2.16 G V$

(d) $U = 21.6 G J$ .

Step by step solution

Parameters.

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

(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

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

Substituting in (1) yields

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

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

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

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

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

The electric field flows through the air is given by

$E A = \frac{Q}{\in_{0}} \Rightarrow E = \frac{Q}{A \in_{0}}$ (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

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

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Short Answer

Step by step solution

Parameters.

(a) Calculating the capacitance of the system.

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

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

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

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