Question

the magnitude of a point charge chosan so that the electric field 50 cm away has magnitude 2 N/C​

Answer 1

Explanation:

let magnitude of charge be Q. 

so,magnitude of electric field (E) at a distance r=0.5m away in given by:- 

Er2KQ , which gives: 

Q=kEr2=9×1092×(0.5)2≃5.56×10−11c

 approximately Q=55.6pc