English
Karnataka Board PUCPUC Science 2nd PUC Class 12

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. - Physics

Advertisements
Advertisements

Question

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!)

Short Note

Solution

We do not get an electric shock as we step out of our house because the original equipotential surfaces of open-air change, keeping our body and the ground at the same potential.

shaalaa.com
Equipotential Surfaces
  Is there an error in this question or solution?
Chapter 2: Electrostatic Potential and Capacitance - Exercise [Page 92]

APPEARS IN

NCERT Physics [English] Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 37.1 | Page 92
NCERT Physics [English] Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.36 (a) | Page 91

RELATED QUESTIONS

Define an equipotential surface.


Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.

  1. Identify an equipotential surface of the system.
  2. What is the direction of the electric field at every point on this surface?

Describe schematically the equipotential surfaces corresponding to

(a) a constant electric field in the z-direction,

(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,

(c) a single positive charge at the origin, and

(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.


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 m2. Will he get an electric shock if he touches the metal sheet next morning?


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?


Draw equipotential surfaces:

(1) in the case of a single point charge and

(2) in a constant electric field in Z-direction. Why are the equipotential surfaces about a single charge not equidistant?

(3) Can electric field exist tangential to an equipotential surface? Give reason


What is the geometrical shape of equipotential surfaces due to a single isolated charge?


Why is there no work done in moving a charge from one point to another on an equipotential surface?


Depict the equipotential surfaces for a system of two identical positive point charges placed a distance(d) apart?


Define equipotential surface. 


Find the amount of work done in rotating an electric dipole of dipole moment 3.2 x 10- 8Cm from its position of stable equilibrium to the position of unstable equilibrium in a uniform electric field if intensity 104 N/C.  


An equipotential surface is that surface ______.

If a unit positive charge is taken from one point to another over an equipotential surface, then ______.

Which of the following is NOT the property of equipotential surface?


Equipotential surfaces ______.

  1. are closer in regions of large electric fields compared to regions of lower electric fields.
  2. will be more crowded near sharp edges of a conductor.
  3. will be more crowded near regions of large charge densities.
  4. will always be equally spaced.

The work done to move a charge along an equipotential from A to B ______.

  1. cannot be defined as `- int_A^B E.dl`
  2. must be defined as `- int_A^B E.dl`
  3. is zero.
  4. can have a non-zero value.

Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.


Find the equation of the equipotentials for an infinite cylinder of radius r0, carrying charge of linear density λ.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×