A piece of cloud having an area of 25 times 1 | vedclass

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(D) The energy density of an electric field is given by $u = \frac{1}{2} \varepsilon_0 E^2$.The electric field $E$ between the cloud and the earth is

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$1475$

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(D) The energy density of an electric field is given by $u = \frac{1}{2} \varepsilon_0 E^2$.The electric field $E$ between the cloud and the earth is $E = \frac{V}{d}$,where $V = 10^5 \, V$ and $d = 0.75 \, km = 750 \, m$.The volume of the space between the cloud and the earth is $Volume = A 	imes d$,where $A = 25 	imes 10^6 \, m^2$.The total energy $U$ is $U = u 	imes Volume = \frac{1}{2} \varepsilon_0 \left( \frac{V}{d} ight)^2 	imes (A 	imes d) = \frac{1}{2} \varepsilon_0 \frac{V^2 A}{d}$.Substituting the values: $U = \frac{1}{2} 	imes (8.85 	imes 10^{-12}) 	imes \frac{(10^5)^2 	imes 25 	imes 10^6}{750}$.$U = \frac{1}{2} 	imes 8.85 	imes 10^{-12} 	imes \frac{10^{10} 	imes 25 	imes 10^6}{750} = \frac{8.85 	imes 25 	imes 10^4}{1500} = \frac{221.25 	imes 10^4}{1500} \approx 1475 \, J$.

$A$ piece of cloud having an area of $25 \times 10^6 \, m^2$ and an electric potential of $10^5 \, V$. If the height of the cloud is $0.75 \, km$,then the energy of the electric field between the earth and the cloud will be.....$J$

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