The force of repulsion between two identical positive charges when kept with a separation 'r' in air is ' ( F ) '. Half the gap between the two charges us filled by a dielectric slab of dielectric constant ( =4 ), Then the new force of repulsion between those two charges becomes

Step-by-step solution

Given, force of repulsion between two identical positive charges with separation, ( r=F ). Half of gap is filled with dielectric constant ( =4 ). Now ( F= 1 4 ₀ q₁ q₂ r² ) where ( q₁ ) and ( q₂ ) are two positive charges. For identical charges, ( q₁=q-2=q ) [ F= 1 4 ₀ q² r² (1) ] When the gap is half filled with dielectric constant then [ F^ = 1 4 ₀ q² (r-t+t k )² ] where ( k ) is dielectric constant ( =4 ; t ) is thickness of dielectric ( =r / 2 ). Therefore, [ array l F^ = 1 4 ₀ q² (r- r 2 + r 2 4 )² = 1 4 ₀ q² (