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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?

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?

Grade:12

1 Answers

Arun
25763 Points
3 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)

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