Lightning transfers charge between a charged cloud and the ground. If the voltage between the two is 1.3x108 V and 3.2 x109 J are transferred, find:A. The total charge moved between the two objectsB. The number of electrons transferred.C. The current, if the lightning strike takes 2.5x10-5 s.

Respuesta :

Given data:

* The value of voltage is,

[tex]V=1.3\times10^8\text{ volts}[/tex]

* The work done in transferring the charge is,

[tex]W=3.2\times10^9\text{ J}[/tex]

* The time taken by the lightning to reach the ground is,

[tex]t=2.5\times10^{-5}\text{ s}[/tex]

Solution:

(a). The voltage between the cloud and ground in terms of the charge transferred is,

[tex]V=\frac{W}{q}[/tex]

where q is the charge transferred,

Substituting the known values,

[tex]\begin{gathered} 1.3\times10^8=\frac{3.2\times10^9}{q} \\ q=\frac{3.2\times10^9}{1.3\times10^8} \\ q=2.46\times10^{9-8} \\ q=24.6\text{ C} \end{gathered}[/tex]

Thus, the amount of charge transferred between the cloud and ground is 24.6 C.

(b). The number of electrons transferred in terms of the net charge and fundamental charge is,

[tex]\begin{gathered} q=ne \\ n=\frac{q}{e} \end{gathered}[/tex]

where e is the fundamental charge,

Substituting the known values,

[tex]\begin{gathered} n=\frac{24.6}{1.6\times10^{-19}} \\ n=15.38\times10^{19} \end{gathered}[/tex]

Thus, the number of electrons transferred is,

[tex]15.38\times10^{19}[/tex]

(c). The current stricks the ground while lightning is,

[tex]I=\frac{q}{t}[/tex]

Substituting the known values,

[tex]\begin{gathered} I=\frac{24.6}{2.5\times10^{-5}} \\ I=9.84\times10^5\text{ A} \end{gathered}[/tex]

Thus, the current striking the ground is,

[tex]\text{9}.84\times10^5\text{ A}[/tex]