Advertisements
Advertisements
प्रश्न
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!)
Advertisements
उत्तर
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.
संबंधित प्रश्न
Draw a sketch of equipotential surfaces due to a single charge (-q), depicting the electric field lines due to the charge
A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
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?
What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
(Hint: The earth has an electric field of about 100 Vm−1 at its surface in the downward direction, corresponding to a surface charge density = −10−9 C m−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
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?
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.
Consider the following statements and select the correct statement(s).
- Electric field lines are always perpendicular to equipotential surface.
- No two equipotential surfaces can intersect each other.
- Electric field lines are in the direction of tangent to an equipotential surface.
Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately.
- The potential at all the points on an equipotential surface is same.
- Equipotential surfaces never intersect each other.
- Work done in moving a charge from one point to other on an equipotential surface is zero.
The work done to move a charge along an equipotential from A to B ______.
- cannot be defined as `- int_A^B E.dl`
- must be defined as `- int_A^B E.dl`
- is zero.
- can have a non-zero value.
Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.
Draw equipotential surfaces for (i) an electric dipole and (ii) two identical positive charges placed near each other.
Equipotential surfaces are shown in figure. Then the electric field strength will be ______.
