Question

The force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric of constant $4$ is inserted between these two charges. If the thickness of the slab is $\dfrac{r}{2}$, then the force between the charges will become:
$\left( A \right)F$
$\left( B \right)\dfrac{3}{2}F$
$\left( C \right)\dfrac{4}{9}F$
$\left( D \right)\dfrac{F}{2}$

Answer 1

Hint: First write the force acted between the two charged particles when it is placed in vacuum and write the force when a dielectric is inserted between them. Then compare the effective radius and find its reaction with the vacuum radius. Then find the effective force where the total distance is the addition of effective distance and the thickness of the dielectric slab.


Complete step by step answer:
As per the problem there is a force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric of constant $4$ is inserted between these two charges. If the thickness of the slab is $\dfrac{r}{2}$.
We need to calculate the force between the charges when a dielectric is inserted.
We know the force between the two charges when it is placed in vacuum and distance between them is r is represented as,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Now the force when it a dielectric is inserted between these two charges will be,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{Kr{'^2}}}$
Where,
K is the dielectric constant and r’ is the effective thickness of the dielectric slab
Now on comparing the dielectric force with the vacuum force we will get,
${r^2} = Kr{'^2}$
From the problem the dielectric constant is $4$.
Hence on putting we will get,
${r^2} = 4r{'^2}$
$ \Rightarrow r = 2r'$
As the thickness of the dielectric is $\dfrac{r}{2}$
Hence,
$r = 2\dfrac{r}{2}$
Now the force between the two charges after inserting a dielectric of thickness $\dfrac{r}{2}$ .
Here the total distance will be,
${r_{total}} = \dfrac{r}{2} + r$
Total Force equals,
${F_{total}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r_{total}}^2}}$
On putting values we will get,
${F_{total}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{{\left( {\dfrac{r}{2} + r} \right)}^2}}}$
We know $r' = \dfrac{r}{4}$
Then,
${F_{total}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{{\left( {\dfrac{r}{2} + r} \right)}^2}}}$
Now on further solving we will get,
${F_{total}} = \dfrac{4}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{9{r^2}}}$
Which is equals to,
${F_{total}} = \dfrac{4}{9}F$
Therefore the correct option is (C).

Note: Remember that the dielectric constant begins to decrease from the bulk value due to the structural discontinuity as the sample if thickness decreases. Also keep in mind that if we increase or decrease the thickness of the dielectric material then the capacitance will decrease or increase respectively.