In a parallel plate capacitor with air between the plates has a capacitance of 8 pF. If the distance between the plates is reduced by half and space between the plates is filled with a dielectric substance, its capacitance becomes 96 pF. Find the dielectric constant of the dielec

Question

In a parallel plate capacitor with air between the plates has a capacitance of 8 pF. If the distance between the plates is reduced by half and space between the plates is filled with a dielectric substance, its capacitance becomes 96 pF. Find the dielectric constant of the dielectric substance.

**OR**

Two point charges $q_1=4q$ and $q_2=-2q$ are placed at a distance $r$ apart. If 25% of the charge is transferred from $q_1$ to $q_2$, then what will be electrostatic force between $q_1$ and $q_2$.

Accepted Answer

>

> [!TLDR]
> Halving d doubles C; the remaining factor of 6 in C must come from the dielectric constant K.
>
> Initial (air) capacitance: $C_1=\dfrac{\epsilon_0 A}{d}=8	ext{ pF}$.
>
> With distance halved and a dielectric of constant K completely filling the gap: $C_2=\dfrac{K\epsilon_0 A}{d/2}=\dfrac{2K\epsilon_0 A}{d}=2K\,C_1$.
>
> Given $C_2=96$ pF: $2K(8)=96 \Rightarrow 16K=96 \Rightarrow K=6$.
> …

Concept understanding — Capacitance of a Parallel Plate Capacitor

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