The force of repulsion between 2 Point charges placed 16cm apart in vaccum is 7.5×10^-10 B. What will be force between them if they are placed in a medium of dielectric constant k=2.5?
Arushi
8 Years agoGrade 12
1 Answer
Arun
8 Years ago
Dear Arushi r = 1.6 cm F(Vac) = 7.5 * 10–10 N K = 2.5 F(vac)/ F(medium) = ε/ε0 = K Now F(medium) = F(vac)/ K = 3 * 10–10 N Regards Arun (askIITians forum expert)