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Capacitance of a Parallel Plate Capacitor Partly Filled with Dielectric Slab between Plates

Derivation→ Let us consider,

The charge on a parallel-plate capacitor = $q$
The area of parallel-plate = $A$
The distance between the parallel-plate = $d$
The dielectric constant of the slab of a material =$K$
The thickness of the material =$t$
The vacuum (or air) between the plates =$(d-t)$ The surface charge density on the plates= $\sigma$
Capacitance of Parallel Plate Capacitor with Dielectric Slab between Plates
The capacitance of Parallel Plate Capacitor with Dielectric Slab between Plates
The electric field in the air between the plates is

$E_{\circ}=\frac{\sigma}{\epsilon_{\circ}}$

$E_{\circ}=\frac{q}{\epsilon_{\circ}\: A} \qquad(1)$

The electric field in the dielectric material

$E=\frac{q}{\epsilon_{\circ}\: K\: A} \qquad(2)$

The potential difference between the plates

$V=E_{\circ}\left( d-t \right)+E\:t \qquad(3)$

Now substitute the value of $E_{\circ}$ and $E$ in the equation $(3)$, Then we get

$V=\frac{q}{\epsilon_{\circ}\: A} \left( d-t \right)+ \frac{q}{\epsilon_{\circ}\: K\: A} \:t $

$V=\frac{q}{\epsilon_{\circ}\: A} \left[ \left( d-t \right)+ \frac{t}{K} \right] \qquad$

The capacitance of the capacitor is

$C=\frac{q}{V}$

Now substitute the value of electric potential $V$ in the above equation then

$C= \frac{q}{\frac{q}{\epsilon_{\circ}\: A} \left[ \left( d-t \right)+ \frac{t}{K} \right]}$

$C= \frac{\epsilon_{\circ}\: A}{\left[ \left( d-t \right)+ \frac{t}{K} \right]} $

Since $K>1$, the 'effective' distance between the plates becomes less than $d$ and so the capacitance increases.

Special Cases→

  1. When the dielectric slab is completely filled between the parallel plates i.e. $t=d$ then the capacitance between the parallel plates

    $C= \frac{K \: \epsilon_{\circ}\: A}{d}$

  2. When there is a vacuum (or air) between the parallel plates i.e. $t=0$, then the capacitance between the parallel plates

    $C_{\circ}= \frac{ \epsilon_{\circ}\: A}{d}$

  3. When there is a slab of metal whose dielectric constant is infinity (K=∞). If the thickness of the slab is t between the parallel plates, then the capacitance between the parallel plates

    $C= \frac{ \: \epsilon_{\circ}\: A}{d-t}$

  4. When the slabs of dielectric constants $K_{1},K_{2},K_{3},K_{4}...........$ and respective thickness $t_{1},t_{2},t_{3},t_{4},........$ is placed in the entire space between the parallel-plates, then the capacitance between the plates

    $C=\frac{\epsilon_{\circ} A}{d- \left( t_{1}+t_{2}+t_{3}+.... \right)+ \left( \frac{t_{1}}{K_{1}}+\frac{t_{2}}{K_{2}}+\frac{t_{3}}{K_{3}}+....... \right)}$

    But $d=t_{1}+t_{2}+t_{3}+....$

    $C=\frac{\epsilon_{\circ} A}{ \left( \frac{t_{1}}{K_{1}}+\frac{t_{2}}{K_{2}}+\frac{t_{3}}{K_{3}}+..... \right)}$

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