Question

A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is released from rest near the stationary charge and is free to move. Which statement best describes the motion of q after it is released? As it moves farther and farther from Q, its speed will keep increasing. As it moves farther and farther from Q, its speed will decrease. As it moves farther and farther from Q, its acceleration will keep increasing. Its speed will be greatest just after it is released. Its acceleration is zero just after it is released. A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At a point P that is 1.25 cm outside the sheet, the magnitude of the electric field due to the sheet E. If the sheet is now stretched so that its sides have length 2d, what is the magnitude of the electric field at P? E E/2 2E E/4 4E

Answer 1

Answer:

Explanation:

When a second positive point charge q is released from rest near the stationary charge and is free to move , the stationary charge will exert a repulsive force upon it which will go on decreasing as the mobile charge moves away from stationary charge. So it will have decreasing but positive acceleration . So its velocity will go on increasing but with decreasing rate.

So the correct statement is

"As it moves farther and farther from Q, its speed will keep increasing".

The electric field near a charged sheet is proportional to the charge density on the charged surface .As the area has been increased 4 times , area charge density will become 1/4 times , so electric field near it will also reduce to 1/4 times . Hence new electric field will be

E / 4 . Ans .