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Define Equipotential Surface.

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प्रश्न

Define equipotential surface. 

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उत्तर

The surfaces on which no work has to be done in order to move a charge is called equipotential surface. 

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2018-2019 (March) Set 1

संबंधित प्रश्न

Define an equipotential surface.


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?


Statement - 1: For practical purpose, the earth is used as a reference at zero potential in electrical circuits.

Statement - 2: The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by `Q/(4piepsilon_0R)`.


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

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.

Equipotential surfaces are denser (closer together) in regions where:


What is the shape of the equipotential surfaces for a single point charge (+q)?


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 \[k = 9 \times 10^9\ \text{N m}^2/\text{C}^2\])


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