Physics, asked by haroon7879, 1 year ago

The force of repulsion between two identical positive charges when kept with the separation are in air is f half the gap between the two charges if filled by a dielectrib slab of dielectri constant 4

Answers

Answered by abhi178
18

it is given that, force of repulsion between two identical positive charge q_1 and q_2 when kept with separation r in air is f.

e.g., f = \frac{kq_1q_2}{r^2}.....(1)

thickness of slab is filled between the charges , t = r/2

dielectric constant of slab, k = 4

force of repulsion when a slab is filled between charges , F = \frac{kq_1q_2}{(r-t+t\sqrt{k})^2}

= \frac{kq_1q_2}{(r-r/2+r)^2}

= \frac{4kq_1q_2}{9r^2}

from equation (1),

F = 4f/9

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