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प्रश्न
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?
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उत्तर
Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The rise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.
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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.
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!)
Why is there no work done in moving a charge from one point to another on an equipotential surface?
The diagrams below show regions of equipotentials.
(i) |
(ii) |
(iii) |
(iv) |
A positive charge is moved from A to B in each diagram.
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.
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.
If the electric field \[\vec{E}\] had a component along (tangent to) an equipotential surface, what contradiction would arise?
The surface of any conductor in electrostatic equilibrium is an equipotential surface because:
What is the shape of the equipotential surfaces for a single point charge (+q)?
For an infinite line charge (infinite cylinder configuration), the equipotential surfaces are:
In a uniform electric field, the equipotential surfaces take the shape of:
A point charge of \[1\ \mu\text{C}\] produces a potential \[V = \frac{kq}{r}\]. At what distance from the charge does the potential equal 90 V? (Take \[k = 9 \times 10^9\ \text{N m}^2/\text{C}^2\])
