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?
$(a)$ 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?
(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.
$(b)$ 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.
Explore More
Similar Questions
An ellipsoidal cavity is carved within a perfect conductor. $A$ positive charge $q$ is placed at the centre of the cavity. The points $A$ and $B$ are on the cavity surface as shown in the figure. Then
Two identical small spheres carry charges of $Q_1$ and $Q_2$ with $Q_1 >> Q_2$. The charges are $d$ distance apart. The force they exert on one another is $F_1$. The spheres are made to touch one another and then separated to distance $d$ apart. The force they exert on one another now is $F_2$. Then $F_1/F_2$ is
Difficult
View SolutionThe electric field is in the direction of the $x$-axis. The work done to move a charge of $0.2 \ C$ by a distance of $2 \ m$ at an angle of $60^\circ$ with the $x$-axis is $4 \ J$. What is the value of the electric field $E$ in $N/C$?
Easy
View SolutionConsider two point charges of equal magnitude and opposite sign separated by a certain distance. The neutral point due to them
Easy
View SolutionIn $1959$,Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density $N$,which is maintained constant. Let the charge on the proton be: $e_p = -(1 + y)e$,where $e$ is the electronic charge.
$(a)$ Find the critical value of $y$ such that expansion may start.
$(b)$ Show that the velocity of expansion is proportional to the distance from the centre.
$(a)$ Find the critical value of $y$ such that expansion may start.
$(b)$ Show that the velocity of expansion is proportional to the distance from the centre.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo