Posted by maria on .
Two electrostatic point charges of +50.0 μC
and +50.0 μC exert a repulsive force on each
other of 197 N.
What is the distance between the two
charges? The Coulomb constant is 8.98755 ×
109 N · m2/C2.
Answer in units of m.

physics 
pete onoriode,
From electric field force b/w two charges=F=Q1Q2/4pieE¤r (square)
force(f)=197,Q1=50,Q2=50,and 4pie E¤r(square)=8.98755*109Nm/C
:>
197=50*50 /8.98755*109*r
197*8.98755*109*r2=50.0*50.0
r2=192730.912/2500
after division then take d square root of d ans. Given to get ur( r) as d distance.