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प्रश्न
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?
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उत्तर
The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.
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संबंधित प्रश्न
Define an equipotential surface.
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.
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.)
Define equipotential surface.
Draw the equipotential surfaces due to an electric dipole.
Write two important characteristics of equipotential surfaces.
The diagrams below show regions of equipotentials.
(i) |
(ii) |
(iii) |
(iv) |
A positive charge is moved from A to B in each diagram.
- The potential at all the points on an equipotential surface is same.
- Equipotential surfaces never intersect each other.
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- must be defined as `- int_A^B E.dl`
- is zero.
- can have a non-zero value.
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The surface of any conductor in electrostatic equilibrium is an equipotential surface because:
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The relation between the electric field \[\vec{E}\] and the potential gradient is expressed as:
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