Q. Force between two identical charges placed at a distance of $r$ in vacuum is $F$. Now a slab of dielectric 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) $\frac{F}{4}$
Explanation
(d) From Coulomb’s law the force $(F)$ between two charges is
$F = \frac{1}{4\pi\varepsilon_{0}k}\frac{q^{2}}{r^{2}}$
First case $k = 1$
$F = \frac{1}{4\pi\varepsilon_{0}}.\frac{q^{2}}{r^{2}}\ \ \ \ \ \ \ \ \ \ \ldots.(i)$
Second case $k = 4$
$F^{'} = \frac{1}{4\pi\varepsilon_{0} \times 4}\ \frac{q^{2}}{r^{2}}\ \ \ \ldots(ii)\ $
$\Longrightarrow \ \ \frac{\ F}{F'} = 4$
${\ \ F}^{'} = \frac{F}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.