Question

Two like charge of 20 micro coulomb are placed 5cm apart in a medium of dielectricconstant 2.5

calculate force between them.

a)288N b)144N c)546 d)1152N​

Answer 1

Answer:

3

Explanation:

1. 576 N nearest option 3

Answer 2

answer : 576 N

explanation : using Coulomb's force,

F = kq1.q2/r²

we know, force is inversely proportional to dielectric constant. i.e., F ∝ 1/dielectric constant

so, F = kq1.q2/xr² , where x is dielectric constant.

here given, q1 = q2 = 20μC = 20 × 10^-6C

r = 5cm = 0.05 m, k = 9 × 10^9 Nm²/C² and x = 2.5

force between charges , F = (9 × 10^9 × 20 × 10^-6 × 20 × 10^-6)/{2.5 × (0.05)²}

= (9 × 4 × 10^-1)/(2.5 × 25 × 10^-4)

= (36 × 10³)/(2.5 × 25)

= 36 × 4 × 4

= 36 × 16

= 576 N