Answer:
Step-by-step 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 .