Answer:
(a) 8 V, (b) 144000 V, (c) 2 x 10^(-8) C
Explanation:
(a) charge, q = 5 μC , Work, W = 40 x 10-^(-6) J
The electric potential is given by
W = q V
(b)
charge, q = 8 x 10^(-6) C, distance, r = 50 cm = 0.5 m
Let the potential is V.
(c)
Work, W = 8 x 10^(-5) J, Potential difference, V = 4000 V
Let the charge is q.
W= q V
If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.
The resultant of the two forces F1 and F2 is
with direction at
north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be
south-east.
If net charge on one object is doubled, then electric force will also get double. It is because they are directly proportional to each other.
Hope this helps!
