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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!)
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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.
RELATED QUESTIONS
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.
Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.
- Identify an equipotential surface of the system.
- 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.
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.)
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
Depict the equipotential surfaces for a system of two identical positive point charges placed a distance(d) apart?
Define equipotential surface.
Draw the equipotential surfaces due to an electric dipole.
S1 and S2 are the two imaginary surfaces enclosing the charges +q and -q as shown. The electric flux through S1 and S2 are respectively ______.
Assertion: Electric field is discontinuous across the surface of a spherical charged shell.
Reason: Electric potential is continuous across the surface of a spherical charged shell.
- 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.
Equipotential surfaces ______.
Consider a uniform electric field in the ẑ direction. The potential is a constant ______.
- in all space.
- for any x for a given z.
- for any y for a given z.
- on the x-y plane for a given z.
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.
Draw equipotential surfaces for (i) an electric dipole and (ii) two identical positive charges placed near each other.
What is meant by an equipotential surface?
