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Why Does a Phonograph Record Attract Dust Particles Just After It is Cleaned?

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प्रश्न

Why does a phonograph record attract dust particles just after it is cleaned?

एक पंक्ति में उत्तर
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उत्तर

When a phonograph record is cleaned, it develops a charge on its surface due to rubbing. This charge attracts the neutral dust particles due to induction.  

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अध्याय 29: Electric Field and Potential - Short Answers [पृष्ठ ११९]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 29 Electric Field and Potential
Short Answers | Q 5 | पृष्ठ ११९

संबंधित प्रश्न

Consider the situation in the figure. The work done in taking a point charge from P to Ais WA, from P to B is WB and from P to C is WC


A point charge q is rotated along a circle in an electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is 


The electric field and the electric potential at a point are E and V, respectively.  


Electric potential decreases uniformly from 120 V to 80 V, as one moves on the x-axis from x = −1 cm to x = +1 cm. The electric field at the origin 

(a) must be equal to 20 Vcm−1
(b) may be equal to 20 Vcm−1
(c) may be greater than 20 Vcm−1
(d) may be less than 20 Vcm−1 


Which of the following quantities does not depend on the choice of zero potential or zero potential energy?


Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is maximum.

 

A ball of mass 100 g and with a charge of 4.9 × 10−5 C is released from rest in a region where a horizontal electric field of 2.0 × 104 N C−1 exists. (a) Find the resultant force acting on the ball. (b) What will be the path of the ball? (c) Where will the ball be at the end of 2 s?


A block of mass m with a charge q is placed on a smooth horizontal table and is connected to a wall through an unstressed spring of spring constant k, as shown in the figure. A horizontal electric field E, parallel to the spring, is switched on. Find the amplitude of the resulting SHM of the block. 


12 J of work has to be done against an existing electric field to take a charge of 0.01 C from A to B. How much is the potential difference  VB − VA


An electric field of 20 NC−1 exists along the x-axis in space. Calculate the potential difference VB − VA where the points A and B are
(a) A = (0, 0); B = (4 m, 2m)
(b) A = (4 m, 2 m); B = (6 m, 5 m)
(c) A = (0, 0); B = (6 m, 5 m)
Do you find any relation between the answers of parts (a), (b) and (c)?  


An electric field  \[\vec{E}  = ( \vec{i} 20 +  \vec{j} 30)   {NC}^{- 1}\]  exists in space. If the potential at the origin is taken to be zero, find the potential at (2 m, 2 m).

 

Which of the following methods can be used to charge a metal sphere positively without touching it? Select the most appropriate.


The unit of electric field is not equivalent to ______.

For distance far away from centre of dipole the change in magnitude of electric field with change in distance from the centre of dipole is ______.

In general, metallic ropes are suspended on the carriers taking inflammable materials. The reason is ______.


When 1014 electrons are removed from a neutral metal sphere, the charge on the sphere becomes ______.


Five charges, q each are placed at the corners of a regular pentagon of side ‘a’ (Figure).

(a) (i) What will be the electric field at O, the centre of the pentagon?

(ii) What will be the electric field at O if the charge from one of the corners (say A) is removed?

(iii) What will be the electric field at O if the charge q at A is replaced by –q?

(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its corners?


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