Question

The force of repulsion between two point chargers placed 16 cm apart in vaccum is 7.5 × 10–10 N. What will be force between them, if they are placed in a medium of dielectric constant k = 2.5?

Answer 1

Answer:

3 × 10^-10 N

Explanation:

given,

r = 1.6 cm

F vacuum = 7.5 ×10^-10 N

K = 2.5

F vacuum / F medium

= [ (1/4Π€o)(q1q2/r²) ] / [ (1/4Π€)(q1q2/r²)

= € / €o

= K

Therefore F medium = F vacuum / K

= 7.5 × 10^-10 / 2.5

= 3 × 10^-10 N

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