Answer the following:
$(a)$ The top of the atmosphere is at about $400 \; kV$ with respect to the surface of the earth,corresponding to an electric field that decreases with altitude. Near the surface of the earth,the field is about $100 \; Vm^{-1}$. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
$(b)$ $A$ man fixes outside his house one evening a two-metre-high insulating slab carrying on its top a large aluminium sheet of area $1 \; m^2$. Will he get an electric shock if he touches the metal sheet next morning?
$(c)$ The discharging current in the atmosphere due to the small conductivity of air is known to be $1800 \; A$ on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words,what keeps the atmosphere charged?
$(d)$ What are the forms of energy into which the electrical energy of the atmosphere is dissipated during lightning?

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(N/A) We do not get an electric shock because our body and the ground form an equipotential surface. When we step out,our body adjusts to the local potential of the atmosphere,and since we are in contact with the ground,we remain at the same potential as the ground,resulting in zero potential difference across our body.
$(b)$ Yes,the man will get an electric shock. The atmospheric discharging current continuously charges the aluminium sheet. Over time,the sheet reaches a high potential relative to the ground. Touching it provides a discharge path through the body.
$(c)$ The atmosphere is kept charged by the global electrical circuit maintained by thunderstorms and lightning occurring continuously across the globe. These act as batteries that pump negative charge to the earth,balancing the discharging current.
$(d)$ During lightning,the electrical energy of the atmosphere is dissipated into light energy,heat energy,and sound energy.

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