Atmospheric Physics

Electricity and Magnetism Level 5

Treating the Earth and the cloud as a parallel plate capacitor, we can estimate the current of lightning. Suppose the base of a storm cloud is 3km above ground, with a surface area roughly ${10}^{8} {m}^{2}$ . If the dielectric constant of air is approximately 1 and the potential difference of the Earth-cloud system is ${10}^{8} V$ , what would the average current (in Amperes) be if the electric energy is discharged in 0.2 seconds?

Submit the answer to the nearest integer.

The answer is 148.

1 solution

Rubayet Tusher
Mar 25, 2016

If we take electric permittivity to be 8.854*10^(-12) $\frac{C^2}{Nm^2}$ , then the exact answer comes out to be 147.566666666667 Amp which can be rounded off to 148 Amp & NOT 150 Amp . So, you should mention something to clarify this matter, @Steven Zheng .

C= $\frac{(epsilon)A}{d}$

Q=CV= $\frac{(epsilon)AV}{d}$

I = $\frac{Q}{t}$ = $\frac{(epsilon)AV}{dt}$ ..................(1)

Now, for this problem, epsilon =8.854*10^(-12) $\frac{C^2}{Nm^2}$

                                   A=10^8 m^2;            

                                  V=10^8 Volt;             

                                  d=3000m  

                               **&**  t=0.2 sec

Plugging in these values in equation number (1), we get I = 147.5666666666667 Amp & therefore, the answer can be written as 150 Amp approximately.

I typed in 147 then 148 and then 146 and I got it wrong. This question is horrible and incorrect.

William G. - 4 years, 3 months ago

Atmospheric Physics

The answer is 148.

1 solution

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