The force between two identical charges placed at a distance of \( r \) in vacuum is F. Now a sl...
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0The 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 become
(A) \( F \)
(B) \( \frac{3}{5} F \)
(C) \( \frac{4}{9} F \)
(D) \( \frac{F}{4} \)
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