Answer:
The net Electric field at the mid point is 289.19 N/C
Given:
Q = + 71 nC =
Q' = + 42 nC =
Separation distance, d = 1.9 m
Solution:
To find the magnitude of electric field at the mid point,
Electric field at the mid-point due to charge Q is given by:
Now,
Electric field at the mid-point due to charge Q' is given by:
Now,
The net Electric field is given by:
Answer:
(a)-24 ft/s
(b)-16.8 ft/s
(c)-16.16 ft/s
(d) -16 ft/s
Explanation:
The initial velocity, v=48 ft/s
(a)
The average velocity when t=2 and 0.5 seconds later hence t=2.5 is given by
(b)
The average velocity when t lasts 0.05 second then
t=2 and t=2.05 is given by
(c)
The average velocity when t=2 and lasts 0.01 s then t=2.01 then is determined by
(d)
As the intervals get small, that is to imply from 0.05 s later to 0.01 s later as seen in parts a to c, the values are getting small and closer to 16 hence the instantaneous velocity at t=2 is -16 ft/s