Force between two identical charges placed at a distance of `r` in vacuum is `F`.Now a slab of dielectric constant 4 is inserted between these two charges . If the thickness of the slab is `r//2`, then the force between the charges will becomes

Options

A. F
B. `(3)/(5)F`
C. `(4)/(9)F`
D. `(F)/(2)`

Answer

C

Explanation

In vacuum, ` F = (1)/( 4 pi epsilon_(0)) (q^(2))/( r^(2))` . . . (i)
Suppose, force between the chrages is same when charges are r. distance apart in dielectric .
`:. F. = (1)/( 4 pi epsilon_(0)) (q^(2))/( kr .^(2))` . . . (ii)
From (i) and (ii) , `kr.^(2) = r^(2) or, r = sqrt (kr.)`
In the given situation, force between the charges would be
` F. (1)/( 4 pi epsilon_(0)) (q^(2))/((( r)/(2) + sqrt(4) (r)/(2))^(2))= (4)/(9) (q^(2) )/( 4 pi epsilon_(0) r^(2)) = ( 4 F)/( 9)`