Electric Charges And Fields

 

                     Electric Charges and fields

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FORCE AND LAWS OF MOTION, Class IX, Part 3

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                                         In order to understand Newton’s second law of motion, we should first know the meaning of the term momentum of a moving body (or moving object). We know that a cricket ball is much heavier than a tennis ball. Suppose we throw a cricket ball and a tennis ball, both with the same speed or velocity. It will be found that more force is required to stop the cricket ball (Which has more mass) and less force is required to stop the tennis ball (which has less mass). We conclude that the force required to stop a moving body is directly proportional to its mass. If we throw two cricket balls of the same mass at different speeds or velocities, it will be found that more force is required to stop that cricket ball which is moving with higher velocity and less force is required to ...
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FORCE AND LAWS OF MOTION, Class IX, Part 4

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  If a boy wearing roller skates stands facing a wall and pushes the wall with his hands, the boy finds himself moving backward, away from the wall. It appears as if the wall also pushes the boy away. When the boy exerts a force on the wall by pushing it with his hands, then the wall exerts an equal force on the boy in the opposite direction . Since the boy is wearing roller skates, the opposite force exerted by the wall makes him backward, the boy on roller skates is exerting force on the wall towards the left side. The wall exerts an equal force on the boy towards the right side. Due to this, the boy moves in the backward direction. From this discussion, we conclude that when a boy exerts a force on the wall, the wall exerts an equal and opposite force on the boy . This is just an illustration of Newton’s third law of motion. Whenever one body exerts a force on another body, the second body exerts an equal and opposite force on the second body. The force exerted by the firs...
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Electric charges and fields

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Electric Charges and Fields     Problems Question1. The charged particle, of charge 2 μ C and mass 10 milligrams, moving with a velocity of 1000 m / s entres a uniform electric field of strength 10 3 NC -1 directed perpendicular to its direction of motion. Find the velocity and displacement, of the particle after 10 s. Solution. The velocity of the particle, normal to the direction of the field.                   V x = 1000 ms -1 ,       is constant The velocity of the particle, along the direction of the field, after 10 s is given by Question2. Assuming that the charge on an atom is distributed uniformly in a sphere of radius 10 -10 m, what will be the electric field at the surface of the gold atom? For gold, Z = 79. Solution. The charge may be assumed to be concentrated at the center of the sphere of radius 10 -10 m. ...
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