Answer:
Check the explanation
Step-by-step explanation:
given 
 
charge of 
 C 
 
radius 250m 
 
The potential at a point 1m 
 
and 1 V 
 
mass 2 micrograms 
 
charge of +2 Coulombs 
 
the potential is V = K Q / r 
 
V = 9 X 109 X 1.11 X 10-10 / 1 
 
V = 9 X 109 X 1.11 X 10-10 
 
V = 0.999 volt 
 
nearly it is V = 1 v 
 
the energy stored is U = q X V 
 
U = 2 X 1 
 
U = 2 J 
 
the energy is stored in this configuration is U = 2 J